[Leetcode] 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:

基本思路就是进行两次扫描，一次从左往右，一次从右往左。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，复杂度是O(n)。

public class Solution {
    public int candy(int[] ratings) {
        if(ratings.length==0)
            return 0;
        int N=ratings.length;
        int[] candies=new int[N];
        candies[0]=1;
        for(int i=1;i<N;++i){
            if(ratings[i]>ratings[i-1]){
                candies[i]=candies[i-1]+1;
            }else{
                candies[i]=1;
            }
        }
        for(int i=N-2;i>=0;--i){
            if(ratings[i]>ratings[i+1]&&candies[i]<=candies[i+1])
                candies[i]=candies[i+1]+1;
        }
        int result=0;
        for(int iCandy:candies){
            result+=iCandy;
        }
        return result;
    }
}