[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?

https://oj.leetcode.com/problems/candy/

 

思路：扫描两边，第一遍从前往后，如果后一个比前一个大，则糖数增加，否则糖数为1；第二遍扫描，从后向前，如果前一个比后一个大，则前一个值取当前值和右边+1的较大者。

 

public class Solution {
    public int candy(int[] ratings) {
        if (ratings == null || ratings.length == 0)
            return 0;
        int res = 0;
        int len = ratings.length;
        int[] num = new int[len];
        num[0] = 1;

        for (int i = 1; i < len; i++) {
            if (ratings[i] > ratings[i - 1])
                num[i] = num[i - 1] + 1;
            else
                num[i] = 1;

        }

        res += num[len - 1];
        for (int i = len - 2; i >= 0; i--) {
            if (ratings[i] > ratings[i + 1])
                num[i] = Math.max(num[i + 1] + 1, num[i]);
            res += num[i];
        }
        return res;
    }

    public static void main(String[] args) {
        System.out.println(new Solution().candy(new int[] { 4, 2, 3, 4, 1 }));
    }
}

 

第二遍记录：

 

第三遍记录：

　　如果相邻相等的情况，貌似没有规定，就取最小的情况好了。

　　所以第一趟扫描，后续元素比前面大的时候，num增加，否则从1开始。

　　第二趟从右向左更新不满足的情况。

public class Solution {
    public int candy(int[] ratings) {
        int n = ratings.length;
        int[] num = new int[n];
        num[0]=1;
        int res=0;
        
        for(int i=1;i<n;i++){
            if(ratings[i]>ratings[i-1])
                num[i]=num[i-1]+1;
            else
                num[i]=1;        
        }
        
        for(int i=n-2;i>=0;i--){
            if(ratings[i]>ratings[i+1]){
                num[i]=Math.max(num[i],num[i+1]+1);
            }
            res +=num[i];
        }
        res +=num[n-1];
        
        return res;
        
    }
}

 

 

参考：

http://seazenith.iteye.com/blog/1950342

http://www.cnblogs.com/TenosDoIt/p/3389479.html