ZigZag Conversion一、题目如下： The string "PAYPALISHIRING" iswritten in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)P A H N A P L S I I G Y I R And then read line by line: "PAHNAPLSIIGYIR"Write the code that will take a string and make this conversion given a number of rows:string convert(strin...
第k个排列给出集合 [1,2,3,…,n]，其所有元素共有 n! 种排列。按大小顺序列出所有排列情况，并一一标记，当 n = 3 时, 所有排列如下："123""132""213""231""312""321"给定 n 和 k，返回第 k 个排列。说明：给定 n 的范围是 [1, 9]。给定 k 的范围是[1, n!]。示例 1:输入: n = 3, k = 3输出: "213"示例 2: 输入: n = 4, k = 9输出: "2314" 思路：观察一下： 我们取k=17。首先数组都是从0开始计数的，所以k-1=16。从16开始，先可以...
题目：同字符分组难度：Medium题目内容：Given an array of strings, group anagrams together.翻译：给定一组字符串数组，按相同字符组成的字符串分为一组。Example:Input: ["eat", "tea", "tan", "ate", "nat", "bat"], Output: [["ate","eat","tea"],["nat","tan"],["bat"] ]我的思路：因为要分组，那么使用map即可，相同的字符作为键，值为一个List作为对应分组1、首先遍历，每次将当前字符串打断成char数组，然后sort，判断ma...
Patching ArrayGiven a sorted positive integer array nums and an integer n, add/patch elements to the array such that any number in range [1, n] inclusive can be formed by the sum of some elements in the array. Return the minimum number of patches required.Example 1:nums = [1, 3], n = 6Return 1.Combinations of nums are [1], [3], [1,3], which form possible sums of: 1, 3, 4.Now if we add/patch 2 to n...
Add and Search Word - Data structure designDesign a data structure that supports the following two operations:void addWord(word) bool search(word) search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter.For example:addWord("bad") addWord("dad") addWord("mad") search("pad") -> false search("bad") -> true searc...
【082-Remove Duplicates from Sorted List II（排序链表中删除重复元素II）】【LeetCode-面试算法经典-Java实现】【所有题目目录索引】原题 Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. For example, Given 1->2->3->3->4->4->5, return 1->2->5. Given 1->1->1->2->3, return 2->3. 题目大意 给定一个排好序的单链表，删除所有重复...
【JavaScript】Leetcode每日一题-组合总和4【题目描述】给你一个由 不同 整数组成的数组 nums ，和一个目标整数 target 。请你从 nums 中找出并返回总和为 target 的元素组合的个数。题目数据保证答案符合 32 位整数范围。示例1：输入：nums = [1,2,3], target = 4 输出：7 解释： 所有可能的组合为： (1, 1, 1, 1) (1, 1, 2) (1, 2, 1) (1, 3) (2, 1, 1) (2, 2) (3, 1) 请注意，顺序不同的序列被视作不同的组合。 示例2：输入：nu...
A message containing letters from A-Z is being encoded to numbers using the following mapping:‘A‘ -> 1 ‘B‘ -> 2 ... ‘Z‘ -> 26Given a non-empty string containing only digits, determine the total number of ways to decode it.Example 1:Input: "12" Output: 2 Explanation: It could be decoded as "AB" (1 2) or "L" (12). Example 2:Input: "226" Output: 3 Explanation: It could be decoded as "BZ" (2 ...
题目：Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.For example, given the range [5, 7], you should return 4. 思路：开始一位一位做的TLE了，这是网上很简洁很好的一个方法。 很容易理解，因为是2进制，不一样则相与为0，如果第i位一样，i-1位不一样，那么m、n肯定不是相连的，那么其中必然会有一个数字第i位不一样。代码：publicclass Sol...
解题思路分析：该题是在一个整数数组中找到一个和最大的连续子数组，并返回和值。那么如何找到一个和最大的连续子数组呢？我们知道，这肯定需要遍历数组才行；好，那我们就开始遍历数组。首先，我们初始化最大和 sum 和当前和 currSum，对于 currSum，如果它小于0，我们就将数组中下一值赋给它；否则就将数组中下一值与其相加。然后，我们取当前 sum 和 currSum 的最大值即可。代码实现：var maxSubArray = function(nums) {//首先...
题目难度：Easy题目：Given a sorted array, remove the duplicates in-place such that each element appear only once and return the new length.Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.翻译：给定一个排好序的数组后，删除重复的元素，这样每个元素只出现一次，并返回新的长度。不要新建另一个数组分配额外的空间，只能通过修改原有...
题目： Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum. For example: Given the below binary tree and sum = 22,5/ 4 8/ / 11 13 4/ \ / 7 2 5 1return[[5,4,11,2],[5,8,4,5] ]题解： 这道题除了要判断是否有这样的一个path sum，还需要把所有的都可能性结果都返回，所以就用传统的DFS递归解决子问题。代码如下： 1 ...
题目在这里: https://leetcode.com/problems/trapping-rain-water/[标签] Array; Stack; Two Pointers这个题我感觉很难，我自己是完全不会。下面贴的是别人想到的好方法，自己适当做了些简单调整。我觉得，这个解法非常巧妙的使用了双向的 two pointers。用到的变量有左右两个指针，一个在最左边，一个在最右边，然后两个一起往中间走。在走的时候维护两个变量，left barrier 和 right barrier，即左边的挡板和右边的挡板。用例子说...
最先想到的两重for循环，超时，输入的数组很大，一整个页面的数，人直接傻了class Solution {public int numPairsDivisibleBy60(int[] time) {int count = 0;for(int i=0;i<time.length - 1;i++){for(int j=i+1;j<time.length;j++){if((time[i] + time[j]) % 60 == 0){count++;}}}return count;} } 接下来运用余数的思想。一个数除以60的余数为0~59，建立一个数组remainder保存余数出现的次数。 先不考虑余数为0和30的情况。 剩下的...
Given two binary strings, return their sum (also a binary string). For example, a = "11" b = "1" Return "100". 解题思路：JAVA实现如下：static public String addBinary(String a, String b) {if (a.length() < b.length()) {String temp = a;a = b;b = temp;}boolean carry = false;StringBuilder sb = new StringBuilder(a);for (int i = 0; i < b.length(); i++) {if (b.charAt(b.length() - 1 - i) == ‘0‘) {if (sb....