LeetCode - Fruit Into Baskets

本文探讨了一种基于滑动窗口的算法,旨在解决水果篮子问题。通过维护一个包含两种类型水果的窗口,该算法实现了从一系列树木中收集最多数量的水果。文中详细解释了算法流程,包括初始化窗口、更新窗口大小以及如何处理特殊情况。通过多个示例说明了算法的有效性和正确性。
In a row of trees, the i-th tree produces fruit with type tree[i].

You start at any tree of your choice, then repeatedly perform the following steps:

Add one piece of fruit from this tree to your baskets.  If you cannot, stop.
Move to the next tree to the right of the current tree.  If there is no tree to the right, stop.
Note that you do not have any choice after the initial choice of starting tree: you must perform step 1, then step 2, then back to step 1, then step 2, and so on until you stop.

You have two baskets, and each basket can carry any quantity of fruit, but you want each basket to only carry one type of fruit each.

What is the total amount of fruit you can collect with this procedure?

 

Example 1:

Input: [1,2,1]
Output: 3
Explanation: We can collect [1,2,1].
Example 2:

Input: [0,1,2,2]
Output: 3
Explanation: We can collect [1,2,2].
If we started at the first tree, we would only collect [0, 1].
Example 3:

Input: [1,2,3,2,2]
Output: 4
Explanation: We can collect [2,3,2,2].
If we started at the first tree, we would only collect [1, 2].
Example 4:

Input: [3,3,3,1,2,1,1,2,3,3,4]
Output: 5
Explanation: We can collect [1,2,1,1,2].
If we started at the first tree or the eighth tree, we would only collect 4 fruits.
 

Note:

1 <= tree.length <= 40000
0 <= tree[i] < tree.length

用 two points 去维护一个 window, 注意一些coner case

class Solution {
    public int totalFruit(int[] tree) {
        if(tree == null || tree.length == 0){
            return 0;
        }
        int len = tree.length;
        int p = 0;
        int q = 0;
        int max = 0;
        Set<Integer> set = new HashSet<>();
        while(p < len){
                if(set.size() == 0){
                    set.add(tree[p]);
                    p++;
                }
                else if(set.contains(tree[p])){
                    p++;
                }
                else{
                    if(set.size() == 1){
                        set.add(tree[p]);
                        p++;
                    }
                    else{
                        if(p - q > max){
                            max = p - q;
                        }
                        set.clear();
                        
                        int temp = p - 1;
                        int type = tree[temp];
                        set.add(tree[p - 1]);
                        set.add(tree[p]);
                        while(tree[temp] == type){
                            temp --;
                        }
                        q = temp+1;
                    }
                }
        }
        if(p - q > max){
            max = p - q;
        }
        return max;
        
    }
}

 

转载于:https://www.cnblogs.com/incrediblechangshuo/p/9982370.html

内容概要:本文系统阐述了企业新闻发稿在生成式引擎优化(GEO)时代下的全渠道策略与效果评估体系,涵盖当前企业传播面临的预算、资源、内容与效果评估四大挑战,并深入分析2025年新闻发稿行业五大趋势,包括AI驱动的智能化转型、精准化传播、首发内容价值提升、内容资产化及数据可视化。文章重点解析央媒、地方官媒、综合门户和自媒体四类媒体资源的特性、传播优势与发稿策略,提出基于内容适配性、时间节奏、话题设计的策略制定方法,并构建涵盖品牌价值、销售转化与GEO优化的多维评估框架。此外,结合“传声港”工具实操指南,提供AI智能投放、效果监测、自媒体管理与舆情应对的全流程解决方案,并针对科技、消费、B2B、区域品牌四大行业推出定制化发稿方案。; 适合人群:企业市场/公关负责人、品牌传播管理者、数字营销从业者及中小企业决策者,具备一定媒体传播经验并希望提升发稿效率与ROI的专业人士。; 使用场景及目标:①制定科学的新闻发稿策略,实现从“流量思维”向“价值思维”转型;②构建央媒定调、门户扩散、自媒体互动的立体化传播矩阵;③利用AI工具实现精准投放与GEO优化,提升品牌在AI搜索中的权威性与可见性;④通过数据驱动评估体系量化品牌影响力与销售转化效果。; 阅读建议:建议结合文中提供的实操清单、案例分析与工具指南进行系统学习,重点关注媒体适配性策略与GEO评估指标,在实际发稿中分阶段试点“AI+全渠道”组合策略,并定期复盘优化,以实现品牌传播的长期复利效应。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值