Leetcode. 135 Candy(Hard)

**Problem**

There are N children standing in a line. Each child is assigned a rating value. 

You are giving candies to these children subjected to the following requirements:

1.Each child must have at least one candy.

2.Children with a higher rating get more candies than their neighbors.

What is the minimum candies you must give?

**Example**

Input: [0]

Output: 1

**Algorithm**

本题题意为n个小朋友排成一行，每个小朋友拥有一个等级，给每个小朋友分发糖果，要求：

1.每个小朋友都至少有一颗糖；

2.相邻两个小朋友中等级较高者要比较低者拥有更多的糖果。

求最少需要多少糖果。  

本题可采用拓扑排序的思想进行求解。我们先建立有向图：每个小朋友视为顶点，对于每对相邻的小朋友，从等级低的小朋友发出一条边指向等级高的小朋友。记录每个点的度数。然后，将度数为0的点放进队列，更新糖果总数。每次从队列中取出一个点，将该点所指向的点的度数-1，若生成新的度数为0的点，放进队列，更新糖果总数。重复上述过程，即可得到最终糖果数。

建图的时间为O(n)，拓扑排序的时间为O(n)，所以总的时间复杂度为O(n+n)=O(n)

**Code**

#include<iostream>#include<vector>#include<queue>using namespace std;class Solution {public:    int candy(vector<int>& ratings) {        int num = ratings.size();        vector<vector<int> > dag(num);        vector<int> dig(num);        vector<int> can(num, 1);                for(int i = 1; i < num; ++i){        if(ratings[i] < ratings[i-1]){        dag[i].push_back(i-1);        dig[i-1]++;        }        else if(ratings[i] > ratings[i-1]){        dag[i-1].push_back(i);        dig[i]++;        }        }                int ans = 0;        queue<int> q;        for(int i = 0; i < num; ++i){        if(dig[i] == 0){        q.push(i);        ans += can[i];        }        }                while(q.size()){        int temp = q.front();        q.pop();        for(int i=0;i<dag[temp].size();++i){        int cn = dag[temp][i];        dig[cn]--;        can[cn] = max(can[cn], can[temp] + 1);        if(dig[cn] == 0) {        q.push(cn);        ans += can[cn];        }        }        }                return ans;    }};