904. Fruit Into Baskets

yPhantom 2019年10月26日 40次浏览

原题链接

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:

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

Solution

还是窗口法,这题目按照给例子实际上说的是最多能从多少棵树上取水果。

抽象成一般的概念就是,求最长的恰好包含两个重复数字的子数组的长度。

class Solution {
    public int totalFruit(int[] tree) {
        int n = tree.length;
        HashMap<Integer, Integer> count = new HashMap<>();
        int left = 0;
        int res = 0;
        int baskets = 2;
        for (int right = 0; right < n; right++) {
            count.put(tree[right], count.getOrDefault(tree[right], 0) + 1);
            if (count.get(tree[right]) == 1) {
                baskets--;
            }
            while (baskets < 0) {
                count.put(tree[left], count.get(tree[left]) - 1);
                if (count.get(tree[left]) == 0) {
                    baskets++;
                }
                left++;
            }
            res = Math.max(res, right - left + 1);
        }
        return res;
    }
}