LeetCode之Two Sum

题目：

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].

暴力解决方法：

class Solution {public:    vector twoSum(vector& nums, int target) {        bool flag = 0;        vector result;        int temp_size = nums.size()-1;// if don't use this variable,just write it in for(;i

注意两点：

1.如果将写成for(int i=0; i<sums.size()-1;i++),leetcode会报compile error，这时需要一个变量来存sums.size()-1;

2.发现第一个for循环里面如果只遍历到数组倒数第二个数，但是用一个单独变量来存储sums.size()-1，总的运行时间会比循环sums.size()次稍慢。

精简一点版本：

使用unordered_map

class Solution {public:    vector twoSum(vector& nums, int target) {        vector result;        unordered_map findTheRest;        for (int i = 0; i < nums.size(); i++)        {            int theRest = target - nums[i];            if (findTheRest.find(theRest) != findTheRest.end())            {                result.push_back(findTheRest[theRest]);                result.push_back(i);                break;            }            findTheRest[nums[i]] = i;        }        return result;    }};

原理一样，运行更快的版本：

class Solution {public:    vector twoSum(vector& nums, int target) {        unordered_map findTheRest;        for (int i = 0; i < nums.size(); i++)        {            int theRest = target - nums[i];            if (findTheRest.find(theRest) != findTheRest.end())            {                return{findTheRest[theRest],i};            }            findTheRest[nums[i]] = i;        }        return {};    }};