LeetCode(Two Sum)

题目要求：

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

注意数组是没有排序的，输出的索引是原始位置例如 3，2，4 target = 6 输出： 2， 3

思路： 先将原始数组复制到临时数组然后一前一后指针查找， 找到后再在原数组中找到原始位置。

class Solution {public:        vector<int> twoSum(vector<int> &numbers, int target) {        vector<int> tmp(numbers);        sort(tmp.begin(), tmp.end());        int idx1 = 0, idx2 = numbers.size() - 1;        while(idx1 < idx2)        {                        if(target < (tmp[idx1] + tmp[idx2]))                --idx2;            else if(target > (tmp[idx1] + tmp[idx2]))                ++idx1;            else                break;        }        vector<int> ret;        int a = tmp[idx1], b = tmp[idx2];        idx1 = -1, idx2 = -1;        for (size_t i = 0; i < numbers.size(); ++i) {            if (numbers[i] == a || numbers[i] == b) {                if(idx1 == -1)                    idx1 = i + 1;                else                    idx2 = i + 1;            }        }        if(idx1 > idx2)            swap(idx1, idx2);        ret.push_back(idx1);        ret.push_back(idx2);        return ret;    }};

代码：