[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;
}
https://github.com/hfl15