两数之和 Two Sum

问题描述

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.

• mode：easy
• tag： array（数组）

暴力解决

对于数组中每个数，遍历后面的数，看是否有满足条件的。

• 时间复杂度：$O(N^2)$
• 空间复杂度: $O(N)$

Python

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        for i in range(len(nums)):
            j=i+1
            while(j<len(nums)):
                if (nums[j] == target - nums[i]):
                    return [i,j]
                j=j+1

c++

class Solution
{
public:
    vector<int> twoSum(vector<int> &nums, int target)
    {
        int nums_size = nums.size();
        vector<int> res;
        for(int i=0; i<nums_size; i++)
        {
            for(int j=i+1; j<nums_size; j++)
            {
                if(nums[j]==target-nums[i])
                {
                    res.push_back(i);
                    res.push_back(j);                    
                }

            }
        }
        return res;
    } 
};

利用哈希表

对于nums[i]，看target-nums[i]是否在哈希表中，若在，返回两数索引，否则，将nums[i]加入哈希表。

• 时间复杂度: $O(N)$
• 空间复杂度: $O(N)$

Python

利用dict

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        dict={}
        for i in range(len(nums)):
            sub = target - nums[i]
            if sub in dict:
                return [dict[sub],i]
            else:
                dict[nums[i]]=i

C++

利用unordered_map

class Solution
{
public:
    vector<int> twoSum(vector<int> &nums, int target)
    {
        //Key是数字，value是该数字的index
        unordered_map<int, int> hash;
        vector<int> result;
        int numsSize = int(nums.size());
        for (int i = 0; i < numsSize; i++)
        {
            int numberToFind = target - nums[i];

            //如果找到numberToFind，就return
            if (hash.find(numberToFind) != hash.end())
            {
                result.push_back(hash[numberToFind]);
                result.push_back(i);
                return result;
            }

            //如果没有找到，把该数字的index放到hash表中
            hash[nums[i]] = i;
        }
        return result;
    }
};