Candy

weixin_33978044

于 2014-02-18 08:33:00 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/RazerLu/p/3553611.html

本文介绍了一个算法用于解决儿童在线分配糖果的问题，确保每个孩子至少得到一颗糖果，并且糖果数量根据评级递增。通过两次遍历解决左右邻居合法性问题。

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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?

两遍遍历，使得左右邻居都合法

public class Solution {
    public int candy(int[] ratings) {
        int len = ratings.length;
        int count = 0;
        int[] candy = new int[len];
        candy[0] = 1;
        for(int i = 1; i < len; i++){
            if(ratings[i] > ratings[i-1]){
                candy[i] = candy[i-1] +1;
            }else{
                candy[i] = 1;
            }
        }
        
        for(int i = len-2; i>=0; i--){
            if(ratings[i] > ratings[i+1]){
                candy[i] = Math.max(candy[i], candy[i+1]+1);
            }
        }
        
        for(int i = 0; i< candy.length; i++){
            count += candy[i];
        }
        
        return count;
    }
}