GuilinDev

Lc0904

05 August 2008

904 Fruits Into Basket

原题概述

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:

  1. Add one piece of fruit from this tree to your baskets. If you cannot, stop.
  2. 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:

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3

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

Example 2:

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4

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:

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4

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:

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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. ‘1 <= tree.length <= 40000’
  2. ‘0 <= tree[i] < tree.length’

题意和分析

Two Pointers问题，主要的思路是利用两个指针来构造一个滑动窗口，且要求窗口中的元素类型不得超过2种。

代码

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class Solution {
    public int totalFruit(int[] tree) {
        int lastFruit = -1;
        int secondLastFruit = -1;
        int lastFruitCount = 0;
        int currentMax = 0;
        int max = 0;

        for (int fruit : tree) {
            if (fruit == lastFruit || fruit == secondLastFruit) {
                currentMax++;
            } else {
                currentMax = lastFruitCount + 1;// last fruit + new fruit
            }

            if (fruit == lastFruit) {
                lastFruitCount++;
            } else {
                lastFruitCount = 1;
            }

            if (fruit != lastFruit) {
                secondLastFruit = lastFruit;
                lastFruit = fruit;
            }

            max = Math.max(max, currentMax);
        }
        return max;
    }
}
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