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], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

我的daim

/** * Note: The returned array must be malloced, assume caller calls free(). */int* twoSum(int* nums, int numsSize, int target) {    int i,j,mark = 0,minus = 1;    if(target < 0) minus = -1;    int *indices = (int*)malloc(sizeof(int) * 2);    for(i=0;i
    
      minus * target) continue;        for(j=i+1;j != i && j
    

别人的代码，将之前访问过的元素存在数组中，O(n)，虽然很快，但是正式的做法应该是存在Hash表里面。

int* twoSum(int* nums, int numsSize, int target) {    int _[100001] = {0}, *index_plus_one = _ + 50000;    for (int i = 0; i < numsSize; i++) {        int rest = target - nums[i];        if (index_plus_one[rest]) {            int *ans = malloc(sizeof(int) * 2);            ans[0] = i;            ans[1] = index_plus_one[rest] - 1;            return ans;        }        else            index_plus_one[nums[i]] = i + 1;    }    return NULL;}