Leecode Week1

Week1：Two Sum

1. Question

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].

2. Solution

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

思路:利用map将数组元素和下标对应起来，然后用一个循环来得到每个元素所要匹配的值并在map中查找是否存在这个值，若存在就返回两个元素的下标，否则继续查找。这里利用map降低了搜索的复杂度，时间复杂度为O(nlog(n))。需要注意的是，每个元素只能使用一次，所以需要判断是否匹配的是同一个值。