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‎1、两数之和.md

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# 1、两数之和
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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。
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你可以假设每种输入只会对应一个答案。但是,数组中同一个元素在答案里不能重复出现。
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你可以按任意顺序返回答案。
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**示例 1:**
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```
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输入:nums = [2,7,11,15], target = 9
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输出:[0,1]
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解释:因为 nums[0] + nums[1] == 9 ,返回 [0, 1] 。
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```
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**示例 2:**
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```
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输入:nums = [3,2,4], target = 6
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输出:[1,2]
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```
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**示例 3:**
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```
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输入:nums = [3,3], target = 6
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输出:[0,1]
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```
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**提示:**
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2 <= nums.length <= 104
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-109 <= nums[i] <= 109
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-109 <= target <= 109
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**只会存在一个有效答案**
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来源:力扣(LeetCode)
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链接:https://leetcode-cn.com/problems/two-sum
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著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
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**java实现**
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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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int a[] = new int [2];
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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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if(nums[i]+nums[j]==target){
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a[0] = i;
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a[1] = j;
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return a;
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}
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}
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}
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return a;
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}
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}
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```
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c语言实现
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```c
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/**
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* Note: The returned array must be malloced, assume caller calls free().
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*/
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int* twoSum(int* nums, int numsSize, int target, int* returnSize){
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for(int i = 0;i<numsSize;i++){
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for(int j = i+1;j<numsSize;j++){
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if(nums[i]+nums[j]==target){
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int *ret = malloc(sizeof(int)*2);
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ret[0] = i;
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ret[1] = j;
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*returnSize = 2;
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return ret;
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}
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}
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}
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*returnSize = 0;
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return NULL;
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}
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```
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