[LeetCode]Candy, 解题报告

回学校写论文差不多快二周的时间了，总结一句话，在学校真舒服。花了7天时间搞定了毕业论文初稿（之所以这么快肯定跟我这三年的工作量相关了），不过今天导师批复第二章还是需要修改。此外，最近迷上打羽毛球，确实很爽，运动量仅此于篮球，可以场地有限，哎，且打且珍惜吧。

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?

LeetCode上AC题目，还是首先读懂题意，这道题我WA了4次，2次是因为没有搞清楚题目到底需要做什么。

读懂题意可以判定这是一道一维动态规划题目，假设dp[i]代表第i个小孩发的糖果数：

当ratings[i] > ratings[i - 1]，必须比前一个多给一块，dp[i] = dp[i - 1] + 1

当ratings[i] == ratings[i - 1], 两个排名相同，则当前给一块即可，dp[i] = 1

当ratings[i] < ratings[i - 1], 排名比上一个低，应该比上一个少一块，dp[i] = dp[i - 1] - 1; 但是考虑特殊的case，当dp[i - 1] == 1时，这时设置dp[i]为1，然后依次递归，for (int j = i - 1; j >= 0 && ratings[j] > ratings[j + 1]; j --) { dp[j] ++; }

import java.util.Scanner;


public class Candy {
    public static int candy(int[] ratings) {
        if (ratings == null || ratings.length == 0) {
            return 0;
        }

        int[] dp = new int[ratings.length];
        dp[0] = 1;

        for (int i = 1; i < ratings.length; i++) {
            if (ratings[i] > ratings[i - 1]) {
                dp[i] = dp[i - 1] + 1;
            } else if (ratings[i] == ratings[i - 1]) {
                dp[i] = 1;
            } else {
                if (dp[i - 1] > 1) {
                    dp[i] = dp[i - 1] - 1;
                } else {
                    dp[i] = 1;
                    for (int j = i - 1; j >= 0 && ratings[j] > ratings[j + 1]; j --) {
                        dp[j] ++;
                    }
                }
            }
        }

        int res = 0;
        for (int i = 0; i < dp.length; i ++) {
            res += dp[i];
        }
        
        return res;
    }


    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);

        while (cin.hasNext()) {
            int n = cin.nextInt();
            int[] ratings = new int[n];
            for (int i = 0; i < n; i++) {
                ratings[i] = cin.nextInt();
            }

            int res = candy(ratings);

            System.out.println(res);
        }

        cin.close();
    }
}

问题：

不能思维僵化，再次题解题目，题目的要达到的要求是：

• 每个孩子都至少有一个糖果
• 具有较高等级的孩子们比他左右的孩子获得更多的糖果

因此根据题意，思路可以如下：

• 初始化数组dp，数组成员均为1，每个孩子先分配一个糖果
• 从左向右遍历，如果第i个孩子比第i - 1孩子等级高，则dp[i] = dp[i - 1] + 1
• 从右向左遍历，如果第i个孩子比第i + 1孩子等级高并且糖果比i+1糖果少，则dp[i] = dp[i + 1] + 1

这种思路实现的算法复杂度为O(n)

代码如下：

public class Solution {
    public int candy(int[] ratings) {
        if (ratings == null || ratings.length == 0) {
            return 0;
        }

        ArrayList<Integer> dp = new ArrayList<Integer>();
        for (int i = 0; i < ratings.length; i++) {
            dp.add(i, 1);
        }

        for (int i = 1; i < ratings.length; i++) {
            if (ratings[i] > ratings[i - 1]) {
                dp.set(i, dp.get(i - 1) + 1);
            }
        }

        for (int i = ratings.length - 2; i >= 0; i--) {
            if (ratings[i] > ratings[i + 1] && dp.get(i) <= dp.get(i + 1)) {
                dp.set(i, dp.get(i + 1) + 1);
            }
        }

        int res = 0;

        for (int i = 0; i < dp.size(); i++) {
            res += dp.get(i);
        }

        return res;
    }
}