2021-03-20 08:30:29 +00:00
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> 如果阅读时，发现错误，或者动画不可以显示的问题可以添加我微信好友 **[tan45du_one](https://raw.githubusercontent.com/tan45du/tan45du.github.io/master/个人微信.15egrcgqd94w.jpg)** ，备注 github + 题目 + 问题 向我反馈
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>
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> 感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助。
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>
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> 另外希望手机阅读的同学可以来我的 <u>[**公众号：袁厨的算法小屋**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u> 两个平台同步，想要和题友一起刷题，互相监督的同学，可以在我的小屋点击<u>[**刷题小队**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u>进入。
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2021-03-20 07:48:03 +00:00
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#### [1. 两数之和](https://leetcode-cn.com/problems/two-sum/)
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2021-03-17 11:42:49 +00:00
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**题目描述：**
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> 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。
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>
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> 你可以假设每种输入只会对应一个答案。但是，数组中同一个元素不能使用两遍。
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**示例:**
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> 给定 nums = [2, 7, 11, 15], target = 9
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>
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> 因为 nums[0] + nums[1] = 2 + 7 = 9
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> 所以返回 [0, 1]
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题目很容易理解，即让查看数组中有没有两个数的和为目标数，如果有的话则返回两数下标，我们为大家提供两种解法双指针（暴力）法，和哈希表法
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**双指针（暴力）法**
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**解析**
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双指针（L,R）法的思路很简单，L指针用来指向第一个值，R指针用来从第L指针的后面查找数组中是否含有和L指针指向值和为目标值的数。见下图
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例：绿指针指向的值为3，蓝指针需要在绿指针的后面遍历查找是否含有 target - 3 = 2的元素，若含有返回即可。
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**题目代码**
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```java
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class Solution {
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public int[] twoSum(int[] nums, int target) {
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if(nums.length < 2){
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return new int[0];
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}
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int[] rearr = new int[2];
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//查询元素
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for(int i = 0; i < nums.length; i++){
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for(int j = i+1; j < nums.length; j++ ){
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//发现符合条件情况
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if(nums[i] + nums[j] ==target){
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rearr[0] = i;
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rearr[1] = j;
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}
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}
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}
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return rearr;
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}
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}
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```
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**哈希表**
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**解析**
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哈希表的做法很容易理解，我们只需通过一次循环即可，假如我们的 target 值为 9，当前指针指向的值为 2 ，我们只需从哈希表中查找是否含有 7，因为9 - 2 =7 。如果含有 7 我们直接返回即可，如果不含有则将当前的2存入哈希表中，指针移动，指向下一元素。注： key 为元素值，value 为元素索引。
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**动图解析：**
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是不是很容易理解，下面我们来看一下题目代码。
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**题目代码：**
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2021-04-26 06:44:47 +00:00
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Java Code:
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2021-03-17 11:42:49 +00:00
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```java
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class Solution {
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public int[] twoSum(int[] nums, int target) {
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HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
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for(int i = 0; i < nums.length; i++){
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//如果存在则返回
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if(map.containsKey(target-nums[i])){
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return new int[]{map.get(target-nums[i]),i};
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}
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//不存在则存入
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map.put(nums[i],i);
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}
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return new int[0];
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}
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}
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```
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2021-04-26 06:44:47 +00:00
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C++ Code:
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```cpp
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class Solution {
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public:
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vector<int> twoSum(vector<int>& nums, int target) {
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unordered_map<int, int> m;
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for (int i = 0; i < nums.size(); ++i) {
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int t = target - nums[i];
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if (m.count(t)) return { m[t], i };
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2021-04-26 06:55:44 +00:00
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m[nums[i]] = i;
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2021-04-26 06:44:47 +00:00
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}
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return {};
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}
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};
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```
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JS Code:
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```js
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const twoSum = function (nums, target) {
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const map = new Map();
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for (let i = 0; i < nums.length; i++) {
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const diff = target - nums[i];
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if (map.has(diff)) {
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return [map.get(diff), i];
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}
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map.set(nums[i], i);
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}
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};
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```
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2021-03-20 07:48:03 +00:00
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