title: Leet Code TwoSum
date: 2017-07-08 23:18:54
tags:
- LeetCode
- 算法
categories: 算法

題目：

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].

第一反應(yīng)可以把nums循環(huán)兩次，用n^2的時間復(fù)雜度

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
       int n = numbers.size();
        vector<int> ans;
        for (int i = 0; i < n-1; i++) {
            for (int j = i + 1; j < n; j++) {
                if (numbers[i] + numbers[j] == target) {
                    ans.push_back(i);
                    ans.push_back(j);
                    return ans;
                }
            }
        }
        return ans;
    }
};

后來想到可以把每個數(shù)字放到Map里，總的時間復(fù)雜度可以到n。

class Solution{
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        map<int, int> mymap;
        int n = numbers.size();
        vector<int> ans;
        for(int i=0;i<n;i++){
            int t = target - numbers[i];
            if(mymap.count(t) > 0){
                ans.push_back(mymap[t]);
                ans.push_back(i);
                return ans;
            }else{
                mymap[numbers[i]] = i;
            }
        }
        return ans;
    }
};

上面程序的運(yùn)行時間是9ms，打敗了54.67% 。

后來看了前排6ms的代碼，發(fā)現(xiàn)只是把map換成了unorder_map,因?yàn)閙ap是紅黑樹實(shí)現(xiàn)的，會根據(jù)鍵的大小排序，查找的時間復(fù)雜度是n，而unorder_map沒有排序，是hash實(shí)現(xiàn)的，查找的時間復(fù)雜度是常數(shù)級，因此會更快。