[array] leetCode-1-Two Sum-Easy

leetCode-1-Two Sum-Easy

descrition

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].

解析

方法 1 ： 2 重循环去检查两个数的和是否等于 target。时间复杂度-O(n^2)，空间复杂度 O(1)

方法 2 : 以空间换时间，使用 hash 表存储已访问过的数，实际上是省去了方法 1 中内层循环的查找时间，时间复杂度 O(n)，空间复杂度 O(n)

注意：题目的假设，输入保证有且只有一个解；返回的是下标。

code

#include <iostream>
#include <vector>
#include <algorithm>
#include <unordered_map>

using namespace std;

class Solution{
public:
vector<int> twoSum(vector<int>& nums, int target){
return twoSumByMap(nums, target);
}

// time-O(n), space-O(n)
vector<int> twoSumByMap(vector<int>& nums, int target){
vector<int> ans;
unordered_map<int, int> hash; // <num, index>
for(int i=0; i<nums.size(); i++){
int another = target - nums[i];
if(hash.find(another) != hash.end()){
// then complexity of unordered_map.find() is
// average case: constant
// worst case: linear in container size
ans.push_back(hash[another]);
ans.push_back(i);
return ans;
}
hash[nums[i]] = i;
}

return ans;
}

};

int main()
{
freopen("in.txt", "r", stdin);
vector<int> nums;
int target;
int cur;
cin >> target;
while(cin >> cur){
nums.push_back(cur);
}
vector<int> ans = Solution().twoSum(nums, target);
if(!ans.empty())
cout << ans[0] << " " << ans[1] << endl;
else
cout << "no answer" << endl;

fclose(stdin);

return 0;
}