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?
思路：
首先初始化每个人一个糖果，然后这个算法需要遍历两遍，第一遍从左向右遍历，如果右边的小盆友的等级高，等加一个糖果，这样保证了一个方向上高等级的糖果多。然后再从右向左遍历一遍，如果相邻两个左边的等级高，而左边的糖果又少的话，则左边糖果数为右边糖果数加一。最后再把所有小盆友的糖果数都加起来返回即可。
代码：
public class Solution {
public int candy(int[] ratings) {
int[] nums = new int[ratings.length];
for (int i=0; i<ratings.length; i++) {
nums[i] = 1;
}
for (int i=0; i<ratings.length - 1; i++) {
if (ratings[i+1] > ratings[i]) {
nums[i+1] = nums[i] + 1;
}
}
for (int i=ratings.length - 1; i>0; i--) {
if (ratings[i-1] > ratings[i]) {
nums[i-1] = Math.max(nums[i] + 1, nums[i - 1]);
}
}
int res = 0;
for (int i=0; i<ratings.length; i++) {
res += nums[i];
}
return res;
}
}