Leetcode--Two Sum

Description

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].

Solution

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

Review

sum = a + b
解法用到了vector<int>,实际就是int类型；
对于unordered_map<int, int> map，
如果需要内部元素自动排序，使用map，不需要排序使用unordered_map，因为我们这里只需要寻找b = （sum - a）是否在数组里，所以不必排序，使用unordered_map，函数find（）用于返回元素在数组中的位置，若map.find(b) != map.end()，说明b存在。

---

Leetcode–Two Sum

---