資工趴趴熊的小天地

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]
Output: 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]

python:

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        # nums = [2,7,11,15], target = 9
        dict={}
        for i in range(len(nums)):
            a=target-nums[i]  #第一次跑a=7(跑12行)
            if a not in dict: #字典沒有7
                dict[nums[i]]=i #加入 {'2':0}
            else:
                return [ dict[a], i]

C++

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        map<int,int>mp;
        for(int i=0;i<nums.size();i++)
        {mp[nums[i]]=i;}
        for(int i=0;i<nums.size();i++)
        {int a = target-nums[i] ;
            if(mp.count(a)&& mp[a]!=i){
                return {i,mp[a]};
            }
        }
        return {};
    }
};