Leetcode1--两数之和 题目分类:数组.哈希表的应用 1. 题目描述 给定一个整数数组 nums 和一个整数目标值 target,请你在该数组中找出 和为目标值 target 的那 两个 整数,并返回它们的数组下标. 你可以假设每种输入只会对应一个答案.但是,数组中同一个元素在答案里不能重复出现. 你可以按任意顺序返回答案.(原题网址:https://leetcode-cn.com/problems/two-sum/) 示例1: 输入:nums = [2,7,11,15], target…

/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */ #define MAX 1000 int Path[MAX]; int index1; int power(int x,int n){ int i,re; if(n==0)return 1; for(i=0,re=1;i<n;++i){…

题目: Given an array of integers, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note t…

package java_net_test; import java.util.HashMap; import java.util.Iterator; import java.util.Map; public class twoSum { public static void main(String[] args) { // TODO Auto-generated method stub int target=9; int[] num={2,7,11,15}; int[]res; for(int…

题目 :Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution.  (Easy) 解法1: Two pointers 拷贝一份,将数组排序,两根指针分别从前后向中间扫描,找到解为止.再遍历原数组寻找下标添加到结果内.…

 问题描写叙述:在一个数组(无序)中高速找出两个数字,使得两个数字之和等于一个给定的值.如果数组中肯定存在至少一组满足要求. <剑指Offer>P214(有序数组) <编程之美>P176 Que:Given an array of integers, find twonumbers such that they add up to a specific target number. The function twoSum should return indices ofthe…

题目描述: 给定一个整数数组 nums 和一个目标值 target,请你在该数组中找出和为目标值的那 两个 整数,并返回他们的数组下标. 你可以假设每种输入只会对应一个答案.但是,你不能重复利用这个数组中同样的元素. 示例: 给定 nums = [2, 7, 11, 15], target = 9 因为 nums[0] + nums[1] = 2 + 7 = 9 所以返回 [0, 1] 方法一: public static int[] twoSum(int[] nums, int target)…

Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15]…

给定一个整数数组和一个目标值,找出数组中和为目标值的两个数. 你可以假设每个输入只对应一种答案,且同样的元素不能被重复利用. 实例: 给定 nums = [2, 7, 11, 15],target = 9 因为 nums[0] + nums[1] = 2 + 7 = 9 所以返回 [0, 1] 1.我的思路: 可以提交但时间复杂度为O(n^2).其中13到20行为了解决数组里有重复元素时,不能用index的情况.如图 class Solution: def twoSum(self, nums,…

public class Solution { public int[] TwoSum(int[] nums, int target) { ]; ; i < nums.Length; i++) { ; j < nums.Length; j++) { if (nums[i] + nums[j] == target) { ary[] = i; ary[] = j; return ary; } } } return ary; } } https://leetcode.com/problems/two…

Write a function that add two numbers A and B. Example Example 1: Input: a = 1, b = 2 Output: 3 Explanation: return the result of a + b. Example 2: Input: a = -1, b = 1 Output: 0 Explanation: return the result of a + b. Challenge Of course you can ju…

  1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = […

Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the sameelement twice. Example: Given nums = [2, 7, 11, 15],…

//定义二维数组int **array = new int*[row_num]; ;i<row_num;i++) { array[i] = new int[col_num]; } vector<vector<)); //删除排好序数组中重复的数据,返回剩余数组的长度 //int A[]: input array//len: the array length//dup: the times that duplicates can be allowed//return the length…

--题目导航见页面左上角的悬浮框#目录导航#-- 相似题型导航 1.1 twosum两数之和  ||  2.2 3Sum三数之和  ||  2.3 3Sum Closest最接近的三数之和 ------------------------------------------------------------------------------------------------------------------------------------- 2.1 Add Two Numbers俩数…

[Problem:1-Two Sum] Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. [Example] Given…

给定一个整数数组和一个目标值,找出数组中和为目标值的两个数. 你可以假设每个输入只对应一种答案,且同样的元素不能被重复利用. 实例: 给定 nums = [2, 7, 11, 15], target = 9 因为 nums[0] + nums[1] = 2 + 7 = 9 所以返回 [0, 1] /** * @param {number[]} nums * @param {number} target * @return {number[]} */ var twoSum = function(n…

最近在家拧三阶魔方,把初级的玩法掌握了,也就是可以还原六个面了,速度不快,但是也很兴奋.三阶魔方的初级玩法按照套路拧就可以了,每一步需要完成的任务,该步骤转动的方法基本都是固定的,而且变化也并不是特别多.只要按照套路多练习,不考虑速度的情况下还原一个三阶魔方还是很容易的. 编程入门和还原魔方其实差不多,最初也是掌握套路后反复的练习,先从一个生手变成一个熟手,然后再去提高.分享一个段子,在知乎上看到的,如下: 陈康肃公尧咨善射,当世无双,公亦以此自矜.尝射于家圃,有卖油翁释担而立,睨之,久而不去.…

1.两个循环 class Solution: def twoSum(self, nums, target): n=len(nums) for i in range(n): for j in range(i+1,n): if (nums[j] == target - nums[i]): return i,j break else: continue 编译通过但耗时太久 2.一个循环 直接看下相加是target数在不在列表中 class Solution: def twoSum(self, nums…

""" Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums =…

1. 两数之和 class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { vector<int> v; for(int i = 0;i < nums.size();i++){ for(int j = i;j < nums.size();j++){ if(i != j && nums[i] + nums[j] == target){ v.pu…

> 简洁易懂讲清原理,讲不清你来打我~ 输入一个数组和一个整数,从数组中找到两个元素和为这个整数,输出下标 > 模拟法 for i找第一个…

HashMap常用方法以及遍历排序方式 常用方法 map.containsKey() map.put() map1.equals(map2) 遍历方式 Iterator<Map.Entry<String, String>> iterator = map.entrySet().iterator(); while (iterator.hasNext()){ System.out.println(iterator.next()); } 排序方式 根据key排序(最好把要排序的设为key)…

这篇文章介绍Leetcode1到10题的解决思路和相关代码. 1. Two sum 问题描述:给定一个整数数组,返回两个数字的索引,使它们加起来等于一个特定的目标. 例子: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. 常规方法:使用双重循环,第一重从左往右固定索引,计算需要查找的结果,第二层循环从固定索引出发依次向右查找第一层计算的结果.时间复杂度\(…

子集和问题:给定一组数和一个值,从这组数中选出若干个数使其和为给定的值.这是个NPC问题. 1.https://leetcode.com/problems/counting-bits/#/solutions 给定一非负Integer num,求[0,num]每个数的二进制形式中1的个数  f[num+1]. 解法:可用最朴素的方法逐个求,但其实有规律: f[i] = f[i / ] + i % 或  f[i] = f[i&(i-)] + ; 2.Single Number:一组Integer类型…

三个问题: 那些内存需要回收? -- 对象是否存活判断 什么时候回收?   --垃圾回收触发条件 如何回收? --垃圾回收算法 垃圾回收应用  -- 理解GC日志.使用垃圾回收命令和工具 1.  判断对象是否存活 引用计数法 可达性分析算法 引用计数法:给对象增添一个计数器,每当被引用一次,计数器数值+1:引用失效则-1:当计数器为0时,该对象不再被使用. 优点:实现简单,判定效率高. 缺点:不能解决对象之间相互循环引用问题. VM参数配置 -Xms100m -Xmx100m  -XX:+Pri…

Leetcode1: Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. 题意:给定一个整数数组,返回两个数下标满足和等于一…

---恢复内容开始--- 对于每次开机avast喊出的“已经检测到危害”实在忍无可忍了(它只能检测到不能根除很气..)于是重装了系统,回到了win10感觉不赖. ================================================================================ leetcode167Two Sum II leetcode209Minimum Size Subarray Sum leetcode216Combination Sum II…

1 案例1 leetcode-----242 给定两个字符串 s 和 t ,编写一个函数来判断 t 是否是 s 的字母异位词. 示例 1: 输入: s = "anagram", t = "nagaram"输出: true python版本 方法1:直接排序 方法二:API map计数 方法三:数组模拟hash ''' 方法1 按照字母序排序 如果一样则是 时间复杂度是nlogN 快排 方法2 数字符串每个字符串字符的个数,也就是使用map来计算{letter:cou…

""" Given the head of a linked list, we repeatedly delete consecutive sequences of nodes that sum to 0 until there are no such sequences. After doing so, return the head of the final linked list. You may return any such answer. (Note that i…