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1. 新增部分题目
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‎java-note-algorithm/src/main/java/com/leosanqing/leetcode/easy/_70_ClimbingStairs.java‎

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‎java-note-algorithm/src/main/java/com/leosanqing/leetcode/easy/list/_876_middle_of_the_linked_list.java‎

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package com.leosanqing.leetcode.easy.list;
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import com.leosanqing.leetcode.medium.list.ListNode;
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/**
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* @Author: rtliu
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* @Date: 2020/7/22 上午9:28
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* @Package: com.leosanqing.leetcode.easy.list
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* @Description: 1
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* Given a non-empty, singly linked list with head node head, return a middle node of linked list.
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* If there are two middle nodes, return the second middle node.
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* Example 1:
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* Input: [1,2,3,4,5]
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* Output: Node 3 from this list (Serialization: [3,4,5]) The returned node has value 3.
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* (The judge's serialization of this node is [3,4,5]).
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* Note that we returned a ListNode object ans, such that: ans.val = 3,
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* ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.
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* Example 2:
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* Input: [1,2,3,4,5,6]
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* Output: Node 4 from this list (Serialization: [4,5,6])
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* Since the list has two middle nodes with values 3 and 4, we return the second one.
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* @Version: 1.0
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*/
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public class _876_middle_of_the_linked_list {
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public static void main(String[] args) {
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ListNode node1 = new ListNode(6);
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ListNode node2 = new ListNode(5, node1);
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ListNode node3 = new ListNode(4, node2);
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ListNode node4 = new ListNode(3,node3);
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ListNode node5 = new ListNode(2, node4);
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middleNode(node5);
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}
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public static ListNode middleNode(ListNode head) {
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ListNode fakeHead = new ListNode();
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fakeHead.next = head;
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ListNode fastNode = fakeHead;
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ListNode slowNode = fastNode;
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while (fastNode != null ) {
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slowNode =slowNode.next;
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fastNode = fastNode.next;
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if(fastNode == null){
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return slowNode;
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}
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fastNode = fastNode.next;
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}
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return slowNode;
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}
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}

‎java-note-algorithm/src/main/java/com/leosanqing/leetcode/easy/num/_7_reverseInteger.java‎

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* @Author: rtliu
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* @Date: 2020/7/3 下午3:19
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* @Package: com.leosanqing.leetcode.easy.num
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* @Description:
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* ` Given a 32-bit signed integer, reverse digits of an integer.
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* ` Example 1:
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* ` Input: 123
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* ` Output: 321
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* ` Example 2:
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* ` Input: -123
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* ` Output: -321
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* ` Example 3:
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* ` Input: 120
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* ` Output: 21
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* ` Note:
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* ` Assume we are dealing with an environment which could only store integers
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* ` within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem,
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* ` assume that your function returns 0 when the reversed integer overflows.
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* @Description: ` Given a 32-bit signed integer, reverse digits of an integer.
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* ` Example 1:
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* ` Input: 123
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* ` Output: 321
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* ` Example 2:
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* ` Input: -123
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* ` Output: -321
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* ` Example 3:
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* ` Input: 120
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* ` Output: 21
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* ` Note:
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* ` Assume we are dealing with an environment which could only store integers
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* ` within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem,
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* ` assume that your function returns 0 when the reversed integer overflows.
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* @Version: 1.0
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*/
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public class _7_reverseInteger {
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public static void main(String[] args) {
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_7_reverseInteger reverseInteger = new _7_reverseInteger();
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System.out.println(reverseInteger.reverse(-1200));
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}
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public int reverse(int x) {
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long res = 0;
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while (x != 0) {
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res *= 10;
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res += x % 10;
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x /= 10;
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}
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return (int)res == res ? (int)res : 0;
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return (int) res == res ? (int) res : 0;
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}
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}

java-note-algorithm/src/main/java/com/leosanqing/leetcode/easy/num/_9_palindromeNumber.java renamed to java-note-algorithm/src/main/java/com/leosanqing/leetcode/easy/num/_9_palindrome_number.java

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* ` Reads 01 from right to left. Therefore it is not a palindrome.
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* @Version: 1.0
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*/
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public class _9_palindromeNumber {
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public class _9_palindrome_number {
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public static void main(String[] args) {
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System.out.println(isPalindrome(12121));
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}
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public static boolean isPalindrome(int x) {
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// 如果小于0 或者大于零 但是以0结尾的都不是
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if (x < 0 || (x !=0 && x % 10 == 0)) {
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if (x < 0 || (x != 0 && x % 10 == 0)) {
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return false;
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}
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int rev = 0;
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// 只需要一半，就可以进行匹配
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while (x > rev) {
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rev = rev *10 + x%10;
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x /=10;
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rev = rev * 10 + x % 10;
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x /= 10;
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}
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return rev == x || x == rev/10;
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return rev == x || x == rev / 10;
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}
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/**
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* 使用字符串
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*
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* @param x
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* @return
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*/
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public static boolean isPalindrome2(int x) {
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// 如果小于0 或者大于零 但是以0结尾的都不是
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if (x < 0 || (x !=0 && x % 10 == 0)) {
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if (x < 0 || (x != 0 && x % 10 == 0)) {
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return false;
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}
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String s = String.valueOf(x);
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for (int i = 0; i < s.length()/2; i++) {
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if (s.charAt(i) != s.charAt(s.length()-i-1)){
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for (int i = 0; i < s.length() / 2; i++) {
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if (s.charAt(i) != s.charAt(s.length() - i - 1)) {
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return false;
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}
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}

‎java-note-algorithm/src/main/java/com/leosanqing/leetcode/medium/_201_bitwiseANDOfNumbersRange.java‎

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* @Author: rtliu
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* @Date: 2020/6/1 上午10:45
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* @Package: medium
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* @Description:
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* Given a range [m, n] where 0 <= m <= n <= 2147483647,
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* return the bitwise AND of all numbers in this range, inclusive.
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*
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* 给定一个范围
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* 返回这个范围内的所有数字的 按位与 的 和
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*
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* Example 1:
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*
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* Input: [5,7]
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* Output: 4
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* Example 2:
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*
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* Input: [0,1]
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* Output: 0
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*
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*
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* @Description: `
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* ` Given a range [m, n] where 0 <= m <= n <= 2147483647,
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* ` return the bitwise AND of all numbers in this range, inclusive.
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* `
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* ` 给定一个范围
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* ` 返回这个范围内的所有数字的 按位与 的 和
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* `
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* ` Example 1:
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* `
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* ` Input: [5,7]
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* ` Output: 4
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* ` Example 2:
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* `
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* ` Input: [0,1]
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* ` Output: 0
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* `
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* `
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* @Version: 1.0
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*/
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public class _201_bitwiseANDOfNumbersRange {
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public static void main(String[] args) {
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System.out.println(solution(5,12));
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System.out.println(solution(5, 12));
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System.out.println(solution(5,12));
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System.out.println(solution(5, 12));
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}
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/**
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* 思路很简单，只要从 m 到 n 的任意一个数字中的一位是0， 那么那一位 不管怎么 & 都是0
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* 比如 [5-12]
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* 5 101
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* 6 110
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* 7 111
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* 8 1000
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* 9 1001
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* 10 1010
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* 11 1011
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* 12 1010
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*
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* 我们看 只要我们找到他的最前面几位是一样的时候，再给他补完后面的位数，就找到了
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*
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* 5 101
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* 6 110
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* 7 111
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* 8 1000
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* 9 1001
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* 10 1010
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* 11 1011
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* 12 1010
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* <p>
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* 我们看 只要我们找到他的最前面几位是一样的时候，再给他补完后面的位数，就找到了
5049
*
5150
* @param m
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* @param n
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* @return
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*/
55-
private static int solution(int m ,int n){
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private static int solution(int m, int n) {
5655
int step = 0;
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while(m != n){
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while (m != n) {
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m >>= 1;
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n >>= 1;
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step++;

‎java-note-algorithm/src/main/java/com/leosanqing/leetcode/medium/array/TaskEvData.java‎

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‎java-note-algorithm/src/main/java/com/leosanqing/leetcode/medium/array/_90_subsetsII.java‎

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package com.leosanqing.leetcode.medium.array;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.List;
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/**
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* @Author: rtliu
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* @Date: 2020/7/22 下午3:00
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* @Package: com.leosanqing.leetcode.medium.array
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* @Description: 1
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* ` Given a collection of integers that might contain duplicates, nums, return all possible subsets (the
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* power set).
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* ` Note: The solution set must not contain duplicate subsets.
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* ` Example:
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* ` Input: [1,2,2]
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* ` Output: [ [2], [1], [1,2,2], [2,2], [1,2], [] ]
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* @Version: 1.0
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*/
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public class _90_subsetsII {
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public static void main(String[] args) {
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subsetsWithDup(new int[]{1, 2, 2});
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}
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public static List<List<Integer>> subsetsWithDup(int[] nums) {
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if (nums == null) {
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return null;
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}
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List<List<Integer>> answer = new ArrayList<>();
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// 先排序
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Arrays.sort(nums);
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for (int i = 0; i <= nums.length; i++) {
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backTrace(answer, new ArrayList<>(), nums, 0, i);
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}
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return answer;
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}
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private static void backTrace(List<List<Integer>> answer, List<Integer> list, int[] nums, int position, int max) {
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if (list.size() == max) {
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answer.add(new ArrayList<>(list));
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// if(!answer.contains(new ArrayList<>(list))){
48+
// }
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return;
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}
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for (int i = position; i < nums.length; i++) {
54+
if (i > position && nums[i] == nums[i - 1]) {
55+
continue;
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}
57+
list.add(nums[i]);
58+
backTrace(answer, list, nums, i + 1, max);
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list.remove(list.size() - 1);
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}
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}
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}

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