题目：
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：
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。
代码如下：
1 public int candy(int[] ratings) {
2 if(ratings==null || ratings.length==0)
3 return 0;
4
5 int[] leftnums = new int[ratings.length];
6 int[] rightnums = new int[ratings.length];
7
8 leftnums[0]=1;
9 for(int i=1;i<ratings.length;i++){
10 if(ratings[i]>ratings[i-1])
11 leftnums[i] = leftnums[i-1]+1;
12 else
13 leftnums[i] = 1;
14 }
15
16 rightnums[ratings.length-1] = leftnums[ratings.length-1];
17 for(int i=ratings.length-2;i>=0;i--){
18 if(ratings[i]>ratings[i+1])
19 rightnums[i] = rightnums[i+1]+1;
20 else
21 rightnums[i] = 1;
22
23 }
24
25 int res = 0;
26 for(int i = 0; i<ratings.length; i++)
27 res += Math.max(leftnums[i],rightnums[i]);
28
29 return res;
30 }
2 if(ratings==null || ratings.length==0)
3 return 0;
4
5 int[] leftnums = new int[ratings.length];
6 int[] rightnums = new int[ratings.length];
7
8 leftnums[0]=1;
9 for(int i=1;i<ratings.length;i++){
10 if(ratings[i]>ratings[i-1])
11 leftnums[i] = leftnums[i-1]+1;
12 else
13 leftnums[i] = 1;
14 }
15
16 rightnums[ratings.length-1] = leftnums[ratings.length-1];
17 for(int i=ratings.length-2;i>=0;i--){
18 if(ratings[i]>ratings[i+1])
19 rightnums[i] = rightnums[i+1]+1;
20 else
21 rightnums[i] = 1;
22
23 }
24
25 int res = 0;
26 for(int i = 0; i<ratings.length; i++)
27 res += Math.max(leftnums[i],rightnums[i]);
28
29 return res;
30 }