LeetCode 两数之和（无序版）

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].

给定一个整数数组，返回两个数字的索引，使它们相加到特定目标。

您可以假设每个输入只有一个解决方案，并且您可能不会两次使用相同的元素。

例：

给定nums = [2,7,11,15]，target = 9， 因为nums [ 0 ] + nums [ 1 ] = 2 + 7 = 9， 返回[ 0，1 ]。

思路

这么简单的题暴力一点啦！

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1.连续遍历nums 更新new_target = target - nums[i];

2.循环遍历找到new_target。

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* twoSum(int* nums, int numsSize, int target) {
    int *ans = (int *)malloc(sizeof(int) * 2);
    for(int i = 0; i < numsSize; i++){
        int new_target = target - nums[i];
        for(int j = i + 1; j < numsSize; j++){
            if(nums[j] == new_target){
                ans[0] = i;
                ans[1] = j;
            }
        }
    }
    return ans;
}