Create 904._Fruit_Into_Baskets.md

docs/Leetcode_Solutions/904._Fruit_Into_Baskets.md

+# 904. Fruit Into Baskets
+
+**<font color=red>难度: Medium</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/contest/weekly-contest-102/problems/fruit-into-baskets/
+
+> 内容描述
+
+```
+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:
+
+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
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(N)******- 空间复杂度: O(N)******
+
+
+
+和第159一样的思路,时间复杂度最差为O(N)，因为有可能整个列表只有两种不同的数字
+
+```python
+class Solution(object):
+    def totalFruit(self, A):
+        """
+        :type tree: List[int]
+        :rtype: int
+        """
+        maps = {}
+        begin, end, counter, res = 0, 0, 0, 0
+        while end < len(A):
+            maps[A[end]] = maps.get(A[end], 0) + 1
+            if maps[A[end]] == 1:
+                counter += 1
+            end += 1  # end points to the next fruit
+            while counter > 2:
+                maps[A[begin]] -= 1
+                if maps[A[begin]] == 0:
+                    counter -= 1
+                begin += 1
+            res = max(res, end - begin)
+        return res
+```