[LeetCode] 1. Two Sum

题：https://leetcode.com/problems/two-sum/description/

题目

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].

思路

1.扫描一遍nums [i] ，将nusm[i ],i 映射到map，时间复杂度为O(logN)，空间复杂度为O(N)；
2.在map中，扫描一遍target - nums [i]，若存在且返回的下标不为i。返回 i 与 上述下标。

笔记

1.map

声明&定义：

map<int,int> tmap;

查找：

map<int,int>::iterator findres = tmap.find(key);
if(findres != tmap.end()) .....(查找到)

findres->first;     //map.key
findres->second;    //map.value

Code

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        int ibegin = 0,iend = nums.size()-1;
        map<int,int> maprec;
        vector<int> res;
        for(int i = 0 ;i<nums.size() ; i++)    maprec[nums[i]] = i;
        for(int i = 0 ;i<nums.size() ; i++){
            int complement = target - nums[i];
            map<int, int>::iterator findres= maprec.find(complement);
            if(findres != maprec.end() && findres->second != i){
                res.push_back(i);
                res.push_back(findres->second);
                return res;
            }
        }
    }
};