LeetCode Candy

原题链接在这里：https://leetcode.com/problems/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?

与Trapping Rain Water相似。每个孩子能拿到多少糖取决于左右两边，和能储存多少水取决于Math.min(左侧挡板最大值, 右侧挡板最大值)一个道理。

所以先用leftNum保存根据左侧邻居，该孩子能拿到多少糖，若是rating比左侧高，就要比左侧多拿一个，否则只需要拿一个。右侧同理。

最后在走一遍，取左右侧中较大值加入res中。

Time O(n), Space O(n).

AC Java:

public class Solution {
    public int candy(int[] ratings) {
        if(ratings == null || ratings.length == 0){
            return 0;
        }
        
        //get how many candies based on left neighbour
        int [] leftNum = new int[ratings.length];
        leftNum[0] = 1;
        for(int i = 1; i<ratings.length; i++){
            if(ratings[i] > ratings[i-1]){
                leftNum[i] = leftNum[i-1]+1;
            }else{
                leftNum[i] = 1;
            }
        }
        
        //get how many candies base on right neighbour
        int [] rightNum = new int[ratings.length];
        rightNum[ratings.length-1] = 1;
        for(int i = ratings.length-2; i>=0; i--){
            if(ratings[i] > ratings[i+1]){
                rightNum[i] = rightNum[i+1]+1;
            }else{
                rightNum[i] = 1;
            }
        }
        
        //total number of candies
        int res = 0;
        for(int i = 0; i<ratings.length; i++){
            res += Math.max(leftNum[i], rightNum[i]);
        }
        return res;
    }
}