【LeetCode】1.Two Sum

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].

给定一个nums数组，和目标值target,要求返回两个数值和为target的数的下标。
将nums中的值和下标通过record翻转过来，然后寻找两个数，不同则输出。
代码如下：

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

自己的笨方法：

暴力求解

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> result(2);
        for(int i=0;i<nums.size();i++){
            for(int j=i+1;j<nums.size();j++){
                if(nums[i]+nums[j]==target){
                    result[0]=i;
                    result[1]=j;
                   
                }
            }
        }
         return result;
    }
};