力扣 1. 两数之和

力扣题目链接

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。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。你可以按任意顺序返回答案。

示例 1：
输入：nums = [2,7,11,15], target = 9
输出：[0,1]

示例 2：
输入：nums = [3,2,4], target = 6
输出：[1,2]

示例 3：
输入：nums = [3,3], target = 6
输出：[0,1]

int* twoSum(int* nums, int numsSize, int target, int* returnSize){
    int *ans = (int *) calloc(2, sizeof(int)); //store these two numbers

    *returnSize = 2;

    for (int i = 0; i < numsSize; ++i) {
        for (int j = i + 1; j < numsSize; ++j) {
            if (nums[i] + nums[j] == target) {
                ans[0] = i;
                ans[1] = j;
                return ans;
            }
        }
    }
    return ans;
}

class Solution {
public:
    vector twoSum(vector& nums, int target) {
        vector ans = {0, 0};
        for (auto i = nums.begin(); i != nums.end(); ++i) {
            for (auto j = i + 1; j != nums.end(); ++j) {
                if ((*i + *j) == target) {
                    ans[0] = distance(nums.begin(), i);
                    ans[1] = distance(nums.begin(), j);
                    return ans;
                }
            }
        }
        return ans;
    }
};

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        nums_len = len(nums)
        for i in range(nums_len):
            for j in range(i + 1, nums_len):
                if nums[i] + nums[j] == target:
                    ans = [i, j]
                    return ans
        return None