leetcode 17 Candy

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:

• Each child must have at least one candy.
• Children with a higher rating get more candies than their neighbors.

What is the minimum candies you must give?

初始化所有小孩糖数目为1，从前往后扫描，如果第i个小孩等级比第i-1个高，那么i的糖数目等于i-1的糖数目+1；

从后往前扫描，如果第i个的小孩的等级比i+1个小孩高,但是糖的数目却小或者相等，那么i的糖数目等于i+1的糖数目+1。

该算法时间复杂度为O(N)

#include<iostream>#include<cstdio>#include<vector>using namespace std;int candy(vector<int> &ratings){        vector<int> count(ratings.size(),1);        int i,len=count.size();        for(i=0;i<len;i++)        {                if(ratings[i+1]>ratings[i]&&count[i+1]<=count[i])                        count[i+1]=count[i]+1;        }        for(i=len-1;i>=0;i--)        {                if(ratings[i-1]>ratings[i]&&count[i-1]<=count[i])                        count[i-1]=count[i]+1;        }        int sum=0;        for(i=0;i<len;i++)                sum+=count[i];        return sum;}int main(int argc,char *argv[]){        freopen("input.txt","r",stdin);        vector<int> v;        int num,tmp;        cin>>num;        while(num--)        {                cin>>tmp;                v.push_back(tmp);        }        cout<<candy(v)<<endl;        return 0;}

这个题只要知道了模拟的方法，就比较简单了。