Leetcode 904 Fruit Into Baskets

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

分析：

  1. 这道题实际上是找到一段最长的距离，使得这段距离内包含的元素只有两种，显然双指针解决，使用一个字典存储已经有的元素及对应的数量。
  2. 对tree[j]分析时，若向字典中加入tree[j]使得字典中元素数量大于2，便开始从左开始删元素，直至字典元素只剩1种，然后将tree[j]加入
  3. 每进行2操作之前需要先比较一次长度，取最大长度作为结果
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class Solution(object):
    def totalFruit(self, tree):
        d = collections.defaultdict(int)
        res = 0
        i,j = 0,0
        while i < len(tree):
            # 将字典大小填充到2
            while j < len(tree) and len(d) < 2: 
                d[tree[j]] += 1
                j += 1
            # 如果继续添加直到字典长度将要变为3的那一刻
            while j < len(tree) and tree[j] in d:
                d[tree[j]] += 1
                j += 1
            # 保存结果
            res = max(res, j-i)
            if j >= len(tree): break
            # 从左删除元素直至字典长度为1
            while len(d) > 1 and i < len(tree):
                d[tree[i]] -= 1
                if d[tree[i]] == 0: del d[tree[i]]
                i += 1
        return res