add

Author: Jack-Lee-Hiter

.idea/workspace.xml

@@ -0,0 +1,34 @@
+# -*- coding:UTF-8 -*-
+import functools
+# Python3.x和Python2.x对于map、reduce、filter的处理变得不同
+# Python3.x中map和filter的输出是一个map型和filter型, 需要从里面取出需要的值
+# Python2.x中map和filter输出的直接是一个list
+# Python3.x中使用reduce需要引入functools
+
+# 使用map把list中的int变为str
+map(str, [1, 2, 3, 4, 5, 6, 7, 8, 9])
+# 使用map()把名字规范化, 首字母大写,其余小写
+def standardName(s):
+    return s.capitalize()
+print([x for x in map(standardName, ['adam', 'LISA', 'barT'])])
+# 在Python2.x中应该使用print(map(standardName, ['adam', 'LISA', 'barT']))
+
+# 使用reduce()输出一个list的所有数的乘积
+def prod(aList):
+    return functools.reduce(lambda x, y: x*y, aList)
+print(prod([1, 2, 3, 4, 5]))
+
+# 使用filter()打印100以内的素数
+def isPrime(n):
+    isPrimeFlag = True
+    if n <= 1:
+        isPrimeFlag = False
+    i = 2
+
+    while i * i <= n:
+        if n % i == 0:
+            isPrimeFlag = False
+            break
+        i += 1
+    return n if isPrimeFlag else None
+print(filter(isPrime, range(101)))

MapReduce_and_filter.py

@@ -0,0 +1,45 @@
+'''
+查找二叉搜索树两个结点的最低公共祖先
+'''
+
+# -*- coding:utf-8 -*-
+class TreeNode:
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+class Solution:
+    def findParent(self, pNode1, pNode2, root):
+        if pNode1 == None or pNode2 == None:
+            return
+        if pNode1 == pNode2:
+            return
+        val1 = pNode1.val
+        val2 = pNode2.val
+        while root != None:
+            if (val1 - root.val) * (val2 - root.val) <= 0:
+                return root.val
+            elif val1 > root.val and val2 > root.val:
+                root = root.right
+            else:
+                root = root.left
+
+        return False
+
+pNode1 = TreeNode(8)
+pNode2 = TreeNode(6)
+pNode3 = TreeNode(10)
+pNode4 = TreeNode(5)
+pNode5 = TreeNode(7)
+pNode6 = TreeNode(9)
+pNode7 = TreeNode(11)
+
+pNode1.left = pNode2
+pNode1.right = pNode3
+pNode2.left = pNode4
+pNode2.right = pNode5
+pNode3.left = pNode6
+pNode3.right = pNode7
+
+S = Solution()
+print(S.findParent(pNode3, pNode7, pNode1))

Target Offer/二叉搜索树的最低公共祖先.py

@@ -1,45 +1,54 @@
-'''
-查找二叉搜索树两个结点的最低公共祖先
-'''
-
-# -*- coding:utf-8 -*-
+# 二叉树的最低公共祖先
+"""
+Definition of TreeNode:
+"""
 class TreeNode:
-    def __init__(self, x):
-        self.val = x
-        self.left = None
-        self.right = None
+    def __init__(self, val):
+        self.val = val
+        self.left, self.right = None, None
 class Solution:
-    def findParent(self, pNode1, pNode2, root):
-        if pNode1 == None or pNode2 == None:
-            return
-        if pNode1 == pNode2:
-            return
-        val1 = pNode1.val
-        val2 = pNode2.val
-        while root != None:
-            if (val1 - root.val) * (val2 - root.val) <= 0:
-                return root.val
-            elif val1 > root.val and val2 > root.val:
-                root = root.right
-            else:
-                root = root.left
+    """
+    @param root: The root of the binary search tree.
+    @param A and B: two nodes in a Binary.
+    @return: Return the least common ancestor(LCA) of the two nodes.
+    """
 
-        return False
-
-pNode1 = TreeNode(8)
-pNode2 = TreeNode(6)
-pNode3 = TreeNode(10)
-pNode4 = TreeNode(5)
-pNode5 = TreeNode(7)
-pNode6 = TreeNode(9)
-pNode7 = TreeNode(11)
+    def lowestCommonAncestor(self, root, A, B):
+        # write your code here
+        if root == None:
+            return False
+        pathA = self.storeNodes(root, A)[0]
+        pathB = self.storeNodes(root, B)[0]
+        if pathA and pathB:
+            lenA, lenB = len(pathA), len(pathB)
+            diff = abs(lenA - lenB)
+            if lenA > lenB:
+                markA = lenA - diff - 1
+                markB = lenB - 1
+            else:
+                markA = lenA - 1
+                markB = lenB - diff - 1
+            while markA >= 0 and markB >= 0:
+                if pathA[markA] == pathB[markB]:
+                    return pathA[markA]
+                markA -= 1
+                markB -= 1
 
-pNode1.left = pNode2
-pNode1.right = pNode3
-pNode2.left = pNode4
-pNode2.right = pNode5
-pNode3.left = pNode6
-pNode3.right = pNode7
+    def storeNodes(self, root, targetNode):
+        if root == None or targetNode == None:
+            return []
+        elif root.val == targetNode.val:
+            return [[targetNode]]
+        stack = []
+        if root.left:
+            stackLeft = self.storeNodes(root.left, targetNode)
+            for i in stackLeft:
+                i.insert(0, root)
+                stack.append(i)
+        if root.right:
+            stackRight = self.storeNodes(root.right, targetNode)
+            for i in stackRight:
+                i.insert(0, root)
+                stack.append(i)
+        return stack
 
-S = Solution()
-print(S.findParent(pNode3, pNode7, pNode1))

Target Offer/二叉树的最低公共祖先.py

@@ -45,6 +45,24 @@ def Find(self, array, target):
             else:
                 return True
         return False
+    # 扩展, 输出数组中target的个数
+    def searchMatrix(self, matrix, target):
+        if matrix == None or len(matrix) == 0:
+            return 0
+        rows = len(matrix)
+        cols = len(matrix[0])
+        row, col = 0, cols - 1
+        count = 0
+        while row <= rows - 1 and col >= 0:
+            if matrix[row][col] > target:
+                col -= 1
+            elif matrix[row][col] < target:
+                row += 1
+            else:
+                count += 1
+                col -= 1
+        return count
+
 
 array = [[1, 2, 8, 9],
          [2, 4, 9, 12],
@@ -53,10 +71,14 @@ def Find(self, array, target):
 array2 = []
 array3 = [['a', 'b', 'c'],
           ['b', 'c', 'd']]
+array4 = [[62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80],[63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81],[64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82],[65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83],[66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84],[67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85]]
+
+
 findtarget = Solution()
 print(findtarget.Find(array, 10))
 print(findtarget.Find(array, 30))
 print(findtarget.Find(array, 13.0))
 print(findtarget.Find(array, ''))
 print(findtarget.Find(array2, 10))
 print(findtarget.Find(array3, 'b'))
+print(findtarget.searchMatrix(array4, 81))

Target Offer/二维数组查找.py

@@ -15,6 +15,7 @@ def sum0(self, n):
     # 利用非0值作两次非运算返回false, 0作两次非运算返回True
     def sumN(self, n):
         fun = {False: self.sum0, True: self.sumN}
+        # 此处的fun[not not n] 不能写作func[not not n-1], 否则测试用例为0的话, 就会无限次迭代
         return n + fun[not not n](n - 1)
 
     def Sum_Solution2(self, n):