【leetcode 1. Two Sum】

题目链接

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].

思路：
1.朴素解法，双层循环 O(n^2)

Code:

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

2 使用map的find() O(nlogn)

Code:

class Solution {
public:
    vector<int> twoSum(vector<int>& nums,int target) {
        map<int, int>mp;
        for(int i=0; i<nums.size(); i++) {
            if(mp.find(target-nums[i])==mp.end()) { ///未在map中找到另一个数
                mp[nums[i]] = i;
            } else { ///找到满足条件的两个数
                vector<int>ans = {mp[target-nums[i]], i};
                return ans;
            }
        }
    }
};