title: Leet Code TwoSum
date: 2017-07-08 23:18:54
tags:
- LeetCode
- 算法
categories: 算法
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].
第一反应可以把nums循环两次，用n^2的时间复杂度
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;
}
};
后来想到可以把每个数字放到Map里，总的时间复杂度可以到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;
}
};
上面程序的运行时间是9ms，打败了54.67% 。
后来看了前排6ms的代码，发现只是把map换成了unorder_map,因为map是红黑树实现的，会根据键的大小排序，查找的时间复杂度是n，而unorder_map没有排序，是hash实现的，查找的时间复杂度是常数级，因此会更快。
