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?

 

Solution:

基本思路就是进行两次扫描，一次从左往右，一次从右往左。第一次扫描的时候维护对于每一个小孩左边所需要最少的糖果数量，存入数组对应元素中，第二次扫描的时候维护右边所需的最少糖果数，并且比较将左边和右边大的糖果数量存入结果数组对应元素中。这样两遍扫描之后就可以得到每一个所需要的最最少糖果量，从而累加得出结果。方法只需要两次扫描，所以时间复杂度是O(2*n)=O(n)。空间上需要一个长度为n的数组，复杂度是O(n)。

1 public class Solution { 2     public int candy(int[] ratings) { 3         if(ratings.length==0) 4             return 0; 5         int N=ratings.length; 6         int[] candies=new int[N]; 7         candies[0]=1; 8         for(int i=1;i
     
      ratings[i-1]){10                 candies[i]=candies[i-1]+1;11             }else{12                 candies[i]=1;13             }14         }15         for(int i=N-2;i>=0;--i){16             if(ratings[i]>ratings[i+1]&&candies[i]<=candies[i+1])17                 candies[i]=candies[i+1]+1;18         }19         int result=0;20         for(int iCandy:candies){21             result+=iCandy;22         }23         return result;24     }25 }