【leetcode】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?

题解：

第一个孩子给一颗糖，然后从左到右遍历rate数组，如果孩子i+1的rate比孩子i的rate高，那么孩子i+1得到的糖果数目比孩子i得到的糖果数目多1；如果孩子i+1的rate比孩子i的rate低，那么就给孩子i+1一颗糖；
从右往左遍历rate数组，如果当前遍历的孩子i的rate比他右边的孩子的rate高，那么他得到的糖果就比他右边的孩子得到的糖果多1.
累加rate中所有的值，得到总的最少糖果数目。

代码如下：

 1 public class Solution {
 2     public int candy(int[] ratings) {
 3         if(ratings.length == 0)
 4             return 0;
 5         
 6         int[] count = new int[ratings.length];
 7         Arrays.fill(count, 1);
 8         
 9         for(int i = 1;i <= ratings.length-1;i++){
10             if(ratings[i]> ratings[i-1] )
11                 count[i] = count[i-1] + 1; 
12         }
13         
14         int sum = 0;
15         for(int i = ratings.length-1;i >= 1;i--){
16             sum += count[i];
17             if(ratings[i-1] > ratings[i] && count[i-1] <= count[i])
18                 count[i-1] = count[i]+ 1; 
19         }
20         
21         return count[0] + sum;
22     }
23 }

在17行的循环，需要判断i-1个孩子当前获得的糖果数目是否真的比第i个孩子的少，如果真的少，才需要+1.例如：

ratings = {4,2,3,4,1}的时候，第一遍遍历得到的count数组是{1,1,2,3,1}，此时从后往前遍历的时候ratings[3] > ratings[4]，但是count[3]已经大于count[4]了，所以不需要更新count[3] = count[4]+1。