id 14, 第三次作业

3rd/homework_1/id_14/103.py

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+"""
+Binary Tree Zigzag Level Order Traversal
+
+Given a binary tree, return the zigzag level order traversal of its nodes' values.
+(ie, from left to right, then right to left for the next level and alternate between).
+
+For example:
+Given binary tree [3,9,20,null,null,15,7],
+    3
+   / \
+  9  20
+    /  \
+   15   7
+return its zigzag level order traversal as:
+[
+  [3],
+  [20,9],
+  [15,7]
+]
+"""
+
+# Definition for a binary tree node.
+class TreeNode(object):
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+from collections import deque
+class Solution(object):
+    def zigzagLevelOrder(self, root):
+        """
+        :type root: TreeNode
+        :rtype: List[List[int]]
+        """
+        if not root:
+            return []
+        queue = deque()
+        queue.append(root)
+        results = []
+        order = 0
+        while len(queue):
+            size = len(queue)
+            currentLevel = []
+            while size > 0:
+                node = queue.popleft()
+                if order:
+                    currentLevel.insert(0, node.val)
+                else:
+                    currentLevel.append(node.val)
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+                size -= 1
+            if len(currentLevel):
+                results.append(currentLevel)
+            order ^= 1
+        return results
+
+root = TreeNode(3)
+root.left = TreeNode(9)
+root.right = TreeNode(20)
+root.right.left = TreeNode(15)
+root.right.right = TreeNode(7)
+
+s = Solution()
+list = s.zigzagLevelOrder(root)
+for l in list:
+    for i in l:
+        print(i)
+    print("...")

3rd/homework_10/id_14/124.py

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+"""
+Binary Tree Maximum Path Sum
+
+Given a binary tree, find the maximum path sum.
+
+For this problem, a path is defined as any sequence of nodes from some
+starting node to any node in the tree along the parent-child connections.
+The path must contain at least one node and does not need to go through the root.
+
+For example:
+Given the below binary tree,
+
+       1
+      / \
+     2   3
+Return 6.
+"""
+# Definition for a binary tree node.
+class TreeNode(object):
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+import sys
+class Solution(object):
+    def __init__(self):
+        self.maxValue = -sys.maxsize
+    def helper(self, root):
+        if not root:
+            return 0
+        left = max(0, self.helper(root.left))
+        right = max(0, self.helper(root.right))
+        self.maxValue = max(self.maxValue, left + right + root.val)
+        return max(left, right) + root.val
+    def maxPathSum(self, root):
+        """
+        :type root: TreeNode
+        :rtype: int
+        """
+        if not root:
+            return 0
+        self.helper(root)
+        return self.maxValue

3rd/homework_11/id_14/113.py

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+"""
+Path Sum II
+
+Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
+
+For example:
+Given the below binary tree and sum = 22,
+              5
+             / \
+            4   8
+           /   / \
+          11  13  4
+         /  \    / \
+        7    2  5   1
+return
+[
+   [5,4,11,2],
+   [5,8,4,5]
+]
+"""
+# 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 helper(self, results, root, sum, path):
+        path.append(root.val)
+        if not root.left and not root.right:
+            if sum == root.val:
+                results.append(path[:])
+        if root.left:
+            self.helper(results, root.left, sum - root.val, path)
+        if root.right:
+            self.helper(results, root.right, sum - root.val, path)
+        path.pop()
+    def pathSum(self, root, sum):
+        """
+        :type root: TreeNode
+        :type sum: int
+        :rtype: List[List[int]]
+        """
+        if not root:
+            return []
+        path = []
+        results = []
+        self.helper(results, root, sum, path)
+        return results
+

3rd/homework_2/id_14/212.py

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+"""
+Word Search II
+
+Given a 2D board and a list of words from the dictionary, find all words in the board.
+
+Each word must be constructed from letters of sequentially adjacent cell, where "adjacent"
+cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.
+
+For example,
+Given words = ["oath","pea","eat","rain"] and board =
+
+[
+  ['o','a','a','n'],
+  ['e','t','a','e'],
+  ['i','h','k','r'],
+  ['i','f','l','v']
+]
+Return ["eat","oath"].
+Note:
+You may assume that all inputs are consist of lowercase letters a-z.
+"""
+
+# 太慢, 被拒了
+class Solution(object):
+    def isExists(self, board, word, y, x, location):
+        if len(word) > len(board) * len(board[0]):
+            return False
+        if location == len(word):
+            return True
+        if y < 0 or x < 0 or y == len(board) or x == len(board[y]):
+            return False
+
+        if board[y][x] != word[location]:
+            return False
+        a = ord('#')
+        board[y][x], a = a, board[y][x]
+        ret = self.isExists(board, word, y + 1, x, location + 1) \
+                or self.isExists(board, word, y - 1, x, location + 1) \
+                or self.isExists(board, word, y, x + 1, location + 1) \
+                or self.isExists(board,word, y, x - 1, location + 1)
+        board[y][x], a = a, board[y][x]
+        return ret
+    def findWords(self, board, words):
+        """
+        :type board: List[List[str]]
+        :type words: List[str]
+        :rtype: List[str]
+        """
+        if not words:
+            return []
+        if not board:
+            return []
+        results = []
+        wordDict = {}
+        for word in words:
+            if word in wordDict:
+                continue
+
+            wordDict[word] = 0
+
+            for y in range(len(board)):
+                if wordDict[word] > 0:
+                    continue
+                for x in range(len(board[0])):
+                    if wordDict[word] > 0:
+                        continue
+                    if self.isExists(board, word, y, x, 0):
+                        wordDict[word] += 1
+                        results.append(word)
+                        if len(results) == len(words):
+                            return results
+        return results
+
+
+board = [
+    ['a'],
+    ['a']
+]
+words = ["a"]
+
+s = Solution()
+print(s.findWords(board, words))

3rd/homework_3/id_14/112.py

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+"""
+Path Sum
+
+Given a binary tree and a sum, determine if the tree has a root-to-leaf path
+such that adding up all the values along the path equals the given sum.
+
+For example:
+Given the below binary tree and sum = 22,
+              5
+             / \
+            4   8
+           /   / \
+          11  13  4
+         /  \      \
+        7    2      1
+return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
+"""
+# 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 hasPathSum(self, root, sum):
+        """
+        :type root: TreeNode
+        :type sum: int
+        :rtype: bool
+        """
+        if not root:
+            return False
+        if root.val == sum and root.left == None and root.right == None:
+            return True
+        return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)
+
+sum = 22
+root = TreeNode(5)
+root.left = TreeNode(4)
+root.right = TreeNode(8)
+root.left.left = TreeNode(11)
+root.left.left.left = TreeNode(7)
+root.left.left.right = TreeNode(2)
+root.right.left = TreeNode(13)
+root.right.right = TreeNode(4)
+root.right.right.right = TreeNode(1)
+
+s = Solution()
+print(s.hasPathSum(root, sum))
+
+

3rd/homework_4/id_14/142.py

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+"""
+Linked List Cycle II
+
+Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
+
+Note: Do not modify the linked list.
+"""
+# Definition for singly-linked list.
+class ListNode(object):
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+class Solution(object):
+    def detectCycle(self, head):
+        """
+        :type head: ListNode
+        :rtype: ListNode
+        """
+        if not head:
+            return None
+        runner = head
+        walker = head
+        while runner and runner.next:
+            walker = walker.next
+            runner = runner.next.next
+            if walker == runner:
+                node = head
+                while node != walker:
+                    walker = walker.next
+                    node = node.next
+                return node
+        return None
+
+class Solution2(object):
+    def detectCycle(self, head):
+        """
+        :type head: ListNode
+        :rtype: ListNode
+        """
+        if not head:
+            return None
+        node = head
+        nodeSet = set()
+        while node:
+            if node in nodeSet:
+                return node
+            nodeSet.add(node)
+            node = node.next
+        return None
+
+head = ListNode(3)
+head.next = ListNode(2)
+node = head.next
+head.next.next = ListNode(0)
+head.next.next.next = ListNode(-4)
+head.next.next.next.next = node
+
+s = Solution()
+l = s.detectCycle(head)
+if l:
+    print(l.val)
+else:
+    print('Null')