问题描述

There are n children standing in a line. Each child is assigned a rating value given in the integer array ratings.

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.

Return the minimum number of candies you need to have to distribute the candies to the children.

测试样例

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Input: ratings = [1,0,2]
Output: 5
Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
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Input: ratings = [1,2,2]
Output: 4
Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
The third child gets 1 candy because it satisfies the above two conditions.

说明

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n == ratings.length
1 <= n <= 2 * 10^4
0 <= ratings[i] <= 2 * 10^4

解题

思路

只需要简单的两次遍历即可：

  1. 把所有孩子的糖果数初始化为 1；

  2. 先从左往右遍历一遍，如果右边孩子的评分比左边的高，则右边孩子的糖果数更新为左边孩子的糖果数加 1；

  3. 再从右往左遍历一遍，如果左边孩子的评分比右边的高，且左边孩子当前的糖果数不大于右边孩子的糖果数，则左边孩子的糖果数更新为右边孩子的糖果数加 1。

通过这两次遍历，分配的糖果就可以满足题目要求了。

这里的贪心策略即为，在每次遍历中，只考虑并更新相邻一侧的大小关系。

补充：

  1. 贪心算法
  2. 数组默认值填充方法 Arrays.fill(array, value)
  3. 数组求和方法 IntStream.of(array).sum()
  4. 时间复杂度 O(n)

代码

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import java.util.Arrays;
import java.util.stream.IntStream;

class Solution {
    public int candy(int[] ratings) {

        // 初始化 candies 数组，表示每个人的糖果
        int[] candies = new int[ratings.length];
        Arrays.fill(candies, 1);    // 最少为 1 个糖果，初始化为 1

        // 从左至右遍历，保证右侧分数高的孩子糖果多 1
        for(int i=0; i<ratings.length-1; ++i) {
            if(ratings[i+1] > ratings[i])
                candies[i+1] = candies[i] + 1;
        }
        // 从右至左遍历，保证左侧分数高，且糖果数小于等于右侧的孩子糖果多 1
        for(int i=ratings.length-1; i>0; --i) {
            if(ratings[i-1] > ratings[i] && candies[i-1] <= candies[i])
                candies[i-1] = candies[i] + 1;
        }
        // candies 数组求和
        return IntStream.of(candies).sum();
    }
}