1. 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.

Example:

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        vector<int> obj;        int v = 0;        for(std::vector<int>::iterator itr1 = nums.begin(); itr1 != nums.end(); ++itr1){            v = target - *itr1;            for(std::vector<int>::iterator itr2 = itr1 + 1; itr2 != nums.end(); ++itr2){                if(*itr2 == v){                    obj.push_back(itr1 - nums.begin() );                    obj.push_back(itr2 - nums.begin() );                }//if            }//for        }//for        return obj;    }};

对nums进行两次遍历，每次从nums的第itr1个位置开始，target与该位置的差值是所需要查找的元素，假设值为v，从第iter1+1处开始对nums进行遍历，查找v，若查找成功，则将将遍历得到的两个元素的下标写入obj中，最后返回；如果最后无法找到这样的两个数，则返回空的obj。