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;
}

作者：fanling999 链接：https://www.cnblogs.com/fanling999/