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关联LeetCode题号135

本题特点

• 贪心
• 两次遍历，一次正序遍历，只比较左边，左边比右边大的情况 i-1 i
• 一次倒序遍历，只比较右边的，右边比左边大 i+1 i

本题思路

class Solution:def candy(self, ratings: List[int]) -> int:candy = [1] * len(ratings)# 右大于左for i in range(1, len(ratings)):if ratings[i] > ratings[i-1]:candy[i] = candy[i-1] + 1# 左大于右for j in range(len(ratings)-2, -1, -1):if ratings[j] > ratings[j+1]:candy[j] = max(candy[j+1]+1, candy[j])print(candy)return sum(candy)

package leetcode;import org.junit.jupiter.api.Test;/*** File Description: Candy* Author: Your Name* Date: 2024/12/24*/
public class Candy_135 {public int candy(int[] ratings) {int[] candy = new int[ratings.length];candy[0] = 1;for (int i=1; i<ratings.length; i++){
//            if (ratings[i-1] < ratings[i]){
//                candy[i] = candy[i-1] + 1;
//            }else{
//                candy[i] = 1;
//            }candy[i] = (ratings[i-1] < ratings[i]) ? candy[i-1]+1 : 1;}for(int j=ratings.length-2; j>=0; j--){if (ratings[j] > ratings[j+1]){candy[j] = Math.max(candy[j], candy[j+1]+1);}}int sum = 0;for(int c: candy){sum += c;}return sum;}@Testpublic void TestCandy(){int[] rating = {1,0,2};System.out.println(candy(rating));}
}