[leetcode]Candy

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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?

算法思路：

双向各扫描一次，第一次正向扫描，寻找递增序列，遇到ratings[i] > ratings[i - 1]，则将candy[i] = candy[i - 1]；

第二次寻找逆向递增序列（递减序列），遇到ratings[i] > ratings[i + 1]，则candy[i] = Math.max(candy[i + 1] + 1,candy[i])

 1         if(ratings == null) return 0;
 2         if(ratings.length < 2) return ratings.length;
 3         int[] candy = new int[ratings.length];
 4         int length = ratings.length;
 5         candy[0] = 1;
 6         for(int i = 1; i < length; i++){
 7             if( ratings[i] > ratings[i - 1]){
 8                 candy[i] = candy[i - 1] + 1;
 9             }else{
10                 candy[i] = 1;
11             }
12         }
13         for(int i = length - 2; i >= 0; i--){
14             if(ratings[i] > ratings[i + 1]){
15                 candy[i] = Math.max(candy[i + 1] + 1,candy[i]);
16             }
17         }
18         int res = 0;
19         for(int i = 0; i < length; res+= candy[i++]);
20         return res;
21     

优化，单边扫描，双指针，扫描过程中找到严格递增和递减序列，递增序列则根据个数，将candy从1开始逐个增加，

递减序列需要注意：首先根据元素个数，逆向从1逐个增加。遇到最大值，则同样candy[i] = Math.max(candy[i + 1] + 1,candy[i]);

实现起来，代码略麻烦.....下次见。。。。

[leetcode]Candy,布布扣,bubuko.com

[leetcode]Candy

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原文地址：http://www.cnblogs.com/huntfor/p/3891324.html