leetcode -- Candy

最新推荐文章于 2024-12-18 23:52:49 发布

转载最新推荐文章于 2024-12-18 23:52:49 发布 · 62 阅读

解决一个关于如何最少地分配糖果的问题，确保评分较高的孩子比邻居获得更多的糖果。采用两次遍历的方法，从左到右和从右到左分别进行，确保满足题目要求的同时达到最小的糖果数量。

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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?

[解题思路]

two pass scan, one pass from left to right, next pass from right to left

when find increment sequence, set candy by k, then increment k

something need to attention, the requirement of this problem is that only children with a higher rating need to get more

candies than their neighors, but if two child has the same rating, we do not need to give more candies

the time complexity is O(n), the space complexity is also O(n)

 1 public class Solution {
 2     public int candy(int[] ratings) {
 3         int N = ratings.length;
 4         if(N == 0){
 5             return N;
 6         }
 7         int[] candy = new int[N];
 8         int sum = N;
 9         for(int i = 0, k = 1; i < N; i++){
10             if(i - 1 >= 0 && ratings[i] > ratings[i - 1]){
11                 candy[i] = Math.max(k ++, candy[i]);
12             } else {
13                 k = 1;
14             }
15         }
16         
17         for(int i = N - 1, k = 1; i >= 0; i--){
18             if(i + 1 < N && ratings[i] > ratings[i + 1]){
19                 candy[i] = Math.max(k ++, candy[i]);
20             } else {
21                 k = 1;
22             }
23         }
24         
25         for(int i = 0; i < N; i++){
26             sum += candy[i];
27         }
28         return sum;
29     }
30 }