算法题之Leetcode分糖果

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;
    }
}