寒假作业

lowestCommonAncestor.py

+# Definition for a binary tree node.
+# class TreeNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution(object):
+    def lowestCommonAncestor(self, root, p, q):
+        """
+        :type root: TreeNode
+        :type p: TreeNode
+        :type q: TreeNode
+        :rtype: TreeNode
+        """
+        if (root == None):
+            return None
+        if ((root.val >= p.val and root.val <= q.val) or (root.val >= q.val and root.val <= p.val)):
+            return root
+        if (root.val < p.val and root.val < q.val):
+            return self.lowestCommonAncestor(root.right, p, q)
+
+        if (root.val > p.val and root.val > q.val):
+            return self.lowestCommonAncestor(root.left, p, q)
+        return None
+
+
+    def lowestCommonAncestor1(self, root, p, q):
+        """
+        :type root: TreeNode
+        :type p: TreeNode
+        :type q: TreeNode
+        :rtype: TreeNode
+        """
+        son_dic = {}
+
+        def get_son_dic(root):
+            if not root:
+                return []
+            else:
+                son_dic[root] = [root] + get_son_dic(root.left) + get_son_dic(root.right)
+                return son_dic[root]
+        result = root
+        while True:
+            if p in son_dic[result.left] and q in son_dic[result.left]:
+                result = result.left
+            elif p in son_dic[result.right] and q in son_dic[result.right]:
+                result = result.right
+            else:
+                return result
+

maxProfit.py

+# 给定一个数组，它的第 i 个元素是一支给定股票第 i 天的价格。
+# 如果你最多只允许完成一笔交易（即买入和卖出一支股票），设计一个算法来计算你所能获取的最大利润。注意你不能在买入股票前卖出股票。
+# 示例 1:输入: [7,1,5,3,6,4]输出: 5解释: 在第 2 天（股票价格 = 1）的时候买入，在第 5 天（股票价格 = 6）的时候卖出，最大利润 = 6-1 = 5 。
+# 注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格。
+# 示例 2:输入: [7,6,4,3,1]输出: 0解释: 在这种情况下, 没有交易完成, 所以最大利润为 0。
+
+
+class Solution(object):
+    def maxProfit(self, prices):
+        """
+        :type prices: List[int]
+        :rtype: int
+        """
+        # 列表个数
+        l = len(prices)
+        # 如果列表个数只有1个，那就不用比了
+        if(l < 2):
+            return 0
+        # 设置初始最小价和最大利润
+        min_price = prices[0]
+        max_Profit = 0
+
+        for i in prices:
+            min_price = min(min_price, i)
+            max_Profit = max(max_Profit, i - min_price)
+        return max_Profit
+
+

mergeTwoLists.py

+# 将两个有序链表合并为一个新的有序链表并返回。
+# 新链表是通过拼接给定的两个链表的所有节点组成的。
+# 示例：输入：1->2->4, 1->3->4输出：1->1->2->3->4->4
+
+# Definition for singly-linked list.
+#  class ListNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+
+    # 递归法
+    def mergeTwoLists(self, l1, l2):
+        """
+        :type l1: ListNode
+        :type l2: ListNode
+        :rtype: ListNode
+        """
+        if l1 == None and l2 == None:
+            return None
+        if l1==None:
+            return l2
+        if l2==None:
+            return l1
+        if l1.val<=l2.val:
+            l1.next=self.mergeTwoLists(l1.next,l2)
+            return l1
+        else:
+            l2.next=self.mergeTwoLists(l1,l2.next)
+            return l2
+
+
+
+
+    # def mergeTwoLists2(self, l1, l2):
+    #     """
+    #     :type l1: ListNode
+    #     :type l2: ListNode
+    #     :rtype: ListNode
+    #     """
+    #     head = ListNode(0)
+    #     first = head
+    #     while l1 != None and l2 != None:
+    #         if l1.val <= l2.val:
+    #             head.next = l1
+    #             l1 = l1.next
+    #         else:
+    #             head.next = l2
+    #             l2 = l2.next
+    #         head = head.next
+    #     if l1 != None:
+    #         head.next = l1
+    #     elif l2 != None:
+    #         head.next = l2
+    #     return first.next