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?

Solution:

进行两次扫描，一次从左往右，一次从右往左。

(这道题用到的思路和Trapping Rain Water是一样的，用动态规划。)

第一次扫描的时候维护对于每一个小孩左边所需要最少的糖果数量，存入数组对应元素中，

第二次扫描的时候维护右边所需的最少糖果数，并且比较将左边和右边大的糖果数量存入结果数组对应元素中。

这样两遍扫描之后就可以得到每一个所需要的最最少糖果量，从而累加得出结果。

方法只需要两次扫描，所以时间复杂度是O(2*n)=O(n)。空间上需要一个长度为n的数组，空间复杂度是O(n)。

public class Solution {
    public int candy(int[] ratings) {
        if(ratings == null || ratings.length == 0)
            return 0;
        
        int[] num = new int[ratings.length];
        num[0] = 1;//at least has 1 candy
        
        for(int i = 1; i < ratings.length; i++) {
            if(ratings[i] > ratings[i - 1]) {
                //if child i has rating higher than i-1, which should 1 bigger than its left neighbour
                num[i] = num[i - 1] + 1;
            }else {
                num[i] = 1;
            }
        }
        
        int res = num[ratings.length - 1];
        
        for(int i = ratings.length - 2; i >= 0; i--) {
            int cur = 1;
            if(ratings[i] > ratings[i + 1]) {
                //if child i has rating higher than its right neighbour, but the candies array did not 
                //represented this situation correctly, then correct it.
                cur = num[i + 1] + 1;
            }
            res += Math.max(cur, num[i]);
            num[i] = cur;
        }
        
        return res;
    }
}