Leetcode1：Two Sum

题目描述

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].

分析

easy难度，两层循环遍历即可，注意vector迭代器的使用。需要了解怎么由迭代器计算出元素下标，这个方法很重要，需要掌握。直接使用当前迭代器减去容器begin迭代器，即可计算出当前迭代器指向元素的下标。

AC代码如下：

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target)
	{
    	int temp = 0;
    	vector<int> result(2);//结果
        for(vector<int>::iterator it = nums.begin(); it < nums.end(); ++it)
        {
        	temp = target - (*it);
        	//第二层遍历从当前迭代器的下一位开始，注意要恢复当前迭代器的位置
        	++it;
        	vector<int>::iterator i = it;
        	--it;
        	for(; i < nums.end(); ++i)
        	{
        		if(temp == (*i))
        		{
        			result[0] = it - nums.begin();//计算下标
        			result[1] = i - nums.begin();
        			return result;
				}
			}
		}
		return result;
    }
};