Candy(Hard) 根据排名分糖果

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.

解题思路:DP

这道题和Trapping water那个是一样的想法，因为无论是水坑还是得到糖的小朋友，影响因素都不只一边，都是左右两边的最小值/最大值来决定的。

所以这道题跟上一道一样，也是左右两边遍历数组。

leftnums数组存从左边遍历，当前小朋友对比其左边小朋友，他能拿到糖的数量；

rightnums数组存从右边遍历，当前小朋友对比其右边小朋友，他能拿到的糖的数量。

最后针对这两个数组，每个小朋友能拿到的糖的数量就是这两个数最大的那个数，求总加和就好了。

public int candy(int[] ratings) {  
        if(ratings==null || ratings.length==0)
            return 0;  

        int[] leftnums = new int[ratings.length];  
        int[] rightnums = new int[ratings.length];

        leftnums[0]=1;  
        for(int i=1;i<ratings.length;i++){  
            if(ratings[i]>ratings[i-1])  
                leftnums[i] = leftnums[i-1]+1;  
            else  
                leftnums[i] = 1;  
        }

        rightnums[ratings.length-1] = leftnums[ratings.length-1];  
        for(int i=ratings.length-2;i>=0;i--){
            if(ratings[i]>ratings[i+1]) 
                rightnums[i] = rightnums[i+1]+1;
            else
                rightnums[i] = 1;

        }

        int res = 0;
        for(int i = 0; i<ratings.length; i++)
            res += Math.max(leftnums[i],rightnums[i]);

        return res;  
}