Candy

Candy

问题：

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?

思路：

　　左到右一遍 右到左一遍 

我的代码：

public class Solution {

    public int candy(int[] ratings) {

        if(ratings==null || ratings.length==0)  return 0;

        if(ratings.length == 1) return 1;

        int len = ratings.length;

        int[] nums = new int[len];

        int k = 1;

        for(int i=1; i<len; i++)

        {

            if(ratings[i] < ratings[i+1])

            {

                nums[i] = k;

                nums[i+1] = k+1;

                k++;

            }

            else

            {

                k = 1;

                nums[i] = Math.max(nums[i], k);

                nums[i+1] = Math.max(nums[i+1], k);

            }

        }

        k = 1;

        for(int i=len-1; i>=1; i--)

        {

            if(ratings[i-1] > ratings[i])

            {

                nums[i] = Math.max(nums[i], k);

                nums[i-1] = Math.max(nums[i-1], k+1);

                k++;

            }

            else

            {

                k = 1;

                nums[i-1] = Math.max(nums[i-1], k);

                nums[i] = Math.max(nums[i], k);

            }

        }

        int rst = 0;

        for(int num : nums) rst += num;

        return rst;

    }

}

View Code

他人代码：

public class Solution {

    public int candy(int[] ratings) {

        if(ratings == null || ratings.length == 0) {

            return 0;

        }



        int[] count = new int[ratings.length];

        Arrays.fill(count, 1);

        int sum = 0;

        for(int i = 1; i < ratings.length; i++) {

            if(ratings[i] > ratings[i - 1]) {

                count[i] = count[i - 1] + 1;

            }

        }



        for(int i = ratings.length - 1; i >= 1; i--) {

            sum += count[i];

            if(ratings[i - 1] > ratings[i] && count[i - 1] <= count[i]) {  // second round has two conditions

                count[i-1] = count[i] + 1;

            }

        }

        sum += count[0];

        return sum;

    }

}

View Code

 

学习之处：

• 一个数组有以下几个特点：上升沿，下降沿，波峰，波谷，和是定值，若数组值的取值范围是0--n则正好对应A[index] = index,可以基于此判断缺少那个数字
• 常用的测试用例 null [0] [1,2] [2,1] [1,2,3] [1,2,1]