Write a function that takes an unsigned integer and returns the number of ’1‘ bits it has (also known as the Hamming weight).For example, the 32-bit integer ’11‘ has binary representation 00000000000000000000000000001011, so the function should return 3.解题思路：JAVA实现如下： public int hammingWeight(int n) {int res = 0;while(n!=0){res += n & 1;n >>>= 1;}return res;} 原文：http://www.cnb...
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).For example, this binary tree is symmetric: 1/ 2 2/ \ / 3 4 4 3But the following is not: 1/ 2 2\ 3 3 解题思路：重载一个isSymmetric判断两棵树是否对称即可，JAVA实现如下： public boolean isSymmetric(TreeNode root) {if(root==null)return true; return isSymmetric(root.left,root.right)...
给定一个排序数组，你需要在原地删除重复出现的元素，使得每个元素只出现一次，返回移除后数组的新长度。不要使用额外的数组空间，你必须在原地修改输入数组并在使用 O(1) 额外空间的条件下完成。示例 1:给定数组 nums = [1,1,2], 函数应该返回新的长度 2, 并且原数组 nums 的前两个元素被修改为 1, 2。 你不需要考虑数组中超出新长度后面的元素。示例 2:给定 nums = [0,0,1,1,1,2,2,3,3,4],函数应该返回新的长度 5, 并且原数组 nu...
题目：Given a set of distinct integers, nums, return all possible subsets.Note:Elements in a subset must be in non-descending order.The solution set must not contain duplicate subsets.For example, If nums = [1,2,3], a solution is:[[3],[1],[2],[1,2,3],[1,3],[2,3],[1,2],[] ]题意：给定一个由不同数字组成的数组nums,返回这个数组中的所有的子集。注意： 1.子集中的元素必须是升序排列 2.最终的结...
Note: This is an extension of House Robber.After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previou...
题目： Given a binary tree, return the preorder traversal of its nodes’ values.For example: Given binary tree {1,#,2,3}, 1 2/3return [1,2,3].Note: Recursive solution is trivial, could you do it iteratively?分析： 先根遍历二叉树，就是说，对二叉树中的每一个节点，先访问该节点，再访问其左子树，最后访问其右子树。用迭代的方式先根遍历二叉树，需要借助栈。具体步骤如下： （1）将根结点入栈 （2）进...
https://leetcode.com/problems/maximum-depth-of-binary-tree/Maximum Depth of Binary TreeGiven a binary tree, find its maximum depth.The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node 1/** 2 * Definition for binary tree3 * public class TreeNode {4 * int val;5 * TreeNode left;6 * TreeNode right;7 * TreeNode(int x) { val...
Given two strings s and t, determine if they are isomorphic.Two strings are isomorphic if the characters in s can be replaced to get t.All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.For example, Given "egg", "add", return true.Given "foo", "bar", retur...
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.解题思路一：之前我们有mergeTwoLists(ListNode l1, ListNode l2)方法，直接调用的话，需要k-1次调用，每次调用都需要产生一个ListNode[]，空间开销很大。如果采用分治的思想，对相邻的两个ListNode进行mergeTwoLists，每次将规模减少一半，直到规模变为2为止，空间开销就会小很多。JAVA实现如下： static public ListNode me...
Given a binary tree, find the maximum path sum.The path may start and end at any node in the tree.For example:Given the below binary tree, 1/ 2 3 Return 6. 对于树，我们可以找到其左右子树中终止于根节点的最大路径值，称为最大半路径值，如果都是正数，则可以把它们以及根节点值相加，如果其中有负数，则舍弃那一边的值（即置为零）。如此可以找到包含根节点的最大路径值。然后选择左右子树的最大半路径...
题目描述：Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -> 4 after calling your function.解题思路：链表的常见操作。代码如下：/*** Definition for singly-linked list.* public class ListNode {* int val;* ...
题目： Implement atoi to convert a string to an integer.Hint: Carefully consider all possible input cases. If you wanta challenge, please do not see below and ask yourself what are the possible input cases. Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front. spoilers alert... click to show...
分析与其说是算法题，不如说是语言特性题。 这题要是对Java的String相关函数掌握的比较熟练，写起来的速度（各种意义上）就会很快。 大致的思路都是一致的，差不到哪里去，无非是枚举长度。值得一提的是，从长到短的枚举顺序要比从短到长优得多。代码class Solution {public String longestCommonPrefix(String[] strs) {if (strs == null || strs.length == 0) {return "";}String prefix = strs[0];for (int i = 1; i < strs.len...
题目描述：Given two strings s and t, determine if they are isomorphic.Two strings are isomorphic if the characters in s can be replaced to get t.All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.For example,Given "egg", "add", return true.Given "foo", "ba...
Combination Sum IIGiven a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.Each number in C may only be used once in the combination.Note:All numbers (including target) will be positive integers.Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).The solution set ...