1. Two Sum.md

1. Two Sum

Problem Description

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Analysis and Code

解法 1

• Time - O(n^2)
• 暴力破解法

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

解法 2

• Time - O(nlogn)
• 利用哈希表達到只需要查詢一次，就可以找到 (Target - num) 是否在哈希表內

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> ans;
        map<int, int> m;
        for(int i = 0; i < nums.size(); i++)
        {
            if(m.find(target - nums[i]) != m.end())    // found key in hashmap
            {
                ans.push_back(i);
                ans.push_back(m[target - nums[i]]);
                return ans;
            }
            else
            {
                m[nums[i]] = i;
            }
        }
        return ans;
    }
};