1.从一串数字中找到相加等于target的两个数。TWO SUM

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.

从一些系列数字中找出两个数相加等于target。只有一个结果，且计算过程中，每个数字元素只能被用两次？？

Example:

Given nums = [2, 7, 11, 15], target = 9,

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

return [0, 1].

——————————————————————————————（写法一）————————————————————————————————————

 

public int[] twoSum(int[] nums, int target) {
    for (int i = 0; i < nums.length; i++) {
        for (int j = i + 1; j < nums.length; j++) {
            if (nums[j] == target - nums[i]) {
                return new int[] { i, j };
            }
        }
    }
    throw new IllegalArgumentException("No two sum solution");
}

————————————————————————————————（写法二）——————————————————————————————————————

class Solution {
    public:
    vector<int> twoSum(vector<int> &nums, int target)
    {
        vector<int> result;
        int N = nums.size();
        for (int i = 0; i < N - 1; i++){
            for (int j = i+1; j < N; j++){
                if (nums[i] + nums[j] == target){
                    result.push_back(i);
                    result.push_back(j);
                    return result;
                }
            }
        }
        return result;
    }
};