1.描述

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].

2.分析

3.代码

C语言版

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

C++版

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 + 1; j < nums.size(); ++j) {
                if (nums[i] + nums[j] == target) {
                    v.push_back(i);
                    v.push_back(j);
                    return v;
                }
            }
        }
        return v;
    }
};