題目

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].

頻度： 5

解題之法

class Solution {
public:

    vector<int> twoSum(vector<int>& nums, int target) {
    
        std::map<int, int> my_map;
        vector<int> result;
        
        for(int i = 0; i < nums.size(); i++) {
        
        int complement = target - nums[i];
        
        //if numberToFind is found in map, return them
        if(my_map.find(complement)!=my_map.end()){
            
            result.push_back(i);
            result.push_back(my_map[complement]);//map 用鍵做下標(biāo)來(lái)尋值
            return result;
        }
        //number was not found. Put it in the map.

        //my_map.insert(i,nums[i]);
        // or say 
        //my_map.insert(pair<int,int>(i,nums[i]));
        //my_map.insert(i, nums[i]);
        my_map.insert(pair<int, int>(nums[i], i));
    }

    return result;
}
};

分析

① 用map來(lái)尋值是常數(shù)級(jí)的，不用遍歷；
② 注意map的尋值和插入方法