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?

Example 1:

Input: [1,0,2]
Output: 5
Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.

Example 2:

Input: [1,2,2]
Output: 4
Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
             The third child gets 1 candy because it satisfies the above two conditions.

思路：

默认的每个孩子首先拿1个。都相同。

然后从左边往右边扫描，如果现在rating的孩子大于左边rating的孩子，则让这个孩子多拿1个。

然后从右边往左边扫描，如果现在rating的孩子大于右边rating的孩子，则让这个孩子拿Math.max(candies[i], candies[i+1]+1).

最后计算和。就是结果。

class Solution {
    public int candy(int[] ratings) {
        if(ratings == null || ratings.length == 0) {
            return 0;
        }
        int n = ratings.length;
        int[] res = new int[n];
        for(int i = 0; i < n; i++) {
            res[i] = 1;
        }
        
        // scan from left to right;
        for(int i = 1; i < n; i++) {
            if(ratings[i - 1] < ratings[i]) {
                res[i] = res[i - 1] + 1;
            }
        }
        
        // scan from right to left;
        for(int i = n - 2; i >= 0; i--) {
            if(ratings[i] > ratings[i + 1]) {
                res[i] = Math.max(res[i], res[i + 1] + 1);
            }
        }
        
        int ret = 0;
        for(int i = 0; i < n; i++) {
            ret += res[i];
        }
        return ret;
    }
}