LeetCode力扣之135. Candy

最新推荐文章于 2024-01-11 14:49:03 发布

Lee_Wei4939

最新推荐文章于 2024-01-11 14:49:03 发布

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本文介绍了一个经典的算法问题——如何根据儿童的评分公平分配最少数量的糖果，确保评分高的儿童获得的糖果数多于邻居。通过两次遍历数组实现，首先从左至右确保每个评分较高的孩子比左侧的孩子多得到糖果，然后从右至左调整确保右侧同样满足条件。

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?

package leetCode;

import java.util.Arrays;

/**
 * Created by lxw, liwei4939@126.com on 2018/3/27.
 */
public class L135_Candy {

    /*从左到右遍历一次，再从右到左遍历一次，具体地，
    * 从左到右遍历，使得右边评分更高的元素比左边至少多于1个糖果
    * 从右到左遍历，使得左边评分更高的元素比右边至少多于1个糖果
    * */
    public int candy(int[] ratings){
        int[] candies = new int[ratings.length];
        Arrays.fill(candies, 1);

        for (int i = 1; i < candies.length; i++){
            if (ratings[i] > ratings[i-1]){
                candies[i] = candies[i-1] + 1;
            }
        }

        for (int i = candies.length-2; i >= 0; i--){
            if (ratings[i] > ratings[i+1]){
                candies[i] = Math.max(candies[i],candies[i+1] +1);
            }
        }

        int sum = 0;
        for (int candy : candies){
            sum += candy;
        }
        return sum;
    }

    
}