LeetCode--------candy

最新推荐文章于 2024-04-06 10:40:33 发布

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最新推荐文章于 2024-04-06 10:40:33 发布

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分类专栏： leetcode 动态规划文章标签： leetcode

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本文介绍了一种解决特定问题的算法——如何在满足条件的情况下，用最少数量的糖果奖励一排孩子。该算法通过两次遍历（从左到右及从右到左）确保每个孩子得到的糖果数既符合其评分也达到全局最小。

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题目描述

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?

class Solution {
public:
    int candy(vector<int> &ratings) {
        int N = ratings.size();
        //dp[i]第i+1人分到糖块数
        //sum[i]前1+1人总共糖块数
        vector<int> dp(N, 0);
        vector<int> sum(N, 0);
        dp[0] = 1;sum[0] = 1;
        for (int i=1; i<N; i++){
            if (ratings[i]>ratings[i-1]){
                dp[i] = dp[i-1]+1;
                sum[i] = sum[i-1]+dp[i];
            }else if(ratings[i-1] == ratings[i] || dp[i-1]>1){//
                dp[i] = 1;
                sum[i] =sum[i-1]+dp[i];
            }else {
                int j;
                dp[i] = dp[i-1];
                for (j=i; j>0 && ratings[j-1]>ratings[j] && dp[j-1]<=dp[j]; j--){
                    dp[j-1]++;
                }
                
                for (int k=1; j < i; j++,k++){
                    sum[j] +=k;
                }
                sum[i] = sum[i-1]+dp[i];
            }
        }
        return sum[N-1];
    }
};

附：左右扫描法

class

Solution {

public:

    int

candy(vector<int>
 &ratings) {

        //题意：N个孩子站成一排，每个孩子分配一个分值。给这些孩子派发糖果，满足如下要求：

        //每个孩子至少一个

        //分值更高的孩子比他的相邻位的孩子获得更多的糖果

        //求至少分发多少糖果？

        int

len=ratings.size();

        if(len==1) return

1;

        int

sum=0;

        vector<int>
 v(len,1);//初始将每个孩子的糖果数都设为1

        //从左向右扫描，保证一个方向上分数更大的糖果更多

        for(int

i=1;i<len;i++){

            if(ratings[i]
 > ratings[i-1])

                v[i]=v[i-1]+1;

        }

        //从右向左扫描，保证另一个方向上分数更大的糖果更多

        for(int

i=len-2;i>=0;i--){

            if(ratings[i]
 > ratings[i+1] && v[i] <= v[i+1])

                v[i]=v[i+1]+1;

        }

        for(int

i=0;i<len;i++){

            sum+=v[i];

        }

        return

sum;

    }

};