LeetCode - Fruit Into Baskets

最新推荐文章于 2024-08-09 21:22:03 发布

转载最新推荐文章于 2024-08-09 21:22:03 发布 · 129 阅读

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原文链接：http://www.cnblogs.com/incrediblechangshuo/p/9982370.html

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