update --52

Author: O2Dyokii

docs/剑指offer/Python/47-求1+2+3+...+n.py

@@ -0,0 +1,18 @@
+# -*- coding:utf-8 -*-
+class Solution:
+    def Sum_Solution(self, n):
+        # write code here
+        return self.sumN(n)
+        
+    def sum0(self, n):
+        return 0
+    
+    def sumN(self,n):
+        fun = {False:self.sum0,True: self.sumN}
+        return n + fun[not not n](n - 1)
+    
+# -*- coding:utf-8 -*-
+class Solution:
+    def Sum_Solution(self, n):
+        # write code here
+        return n and self.Sum_Solution(n - 1) + n

docs/剑指offer/Python/48-不用加减乘除做加法.py

@@ -0,0 +1,9 @@
+# -*- coding:utf-8 -*-
+class Solution:
+    def Add(self, num1, num2):
+        # write code here
+        while num2 != 0:
+            temp = num1 ^ num2
+            num2 = (num1 & num2) << 1
+            num1 = temp & 0xFFFFFFFF
+        return num1 if num1 >> 31 == 0 else num1 - 4294967296

docs/剑指offer/Python/49-把字符串转化成整数.py

@@ -0,0 +1,25 @@
+# -*- coding:utf-8 -*-
+class Solution:
+    def StrToInt(self, s):
+        # write code here
+        flag = False
+        if not s or len(s) < 1:
+            return 0
+        num = []
+        numdict = {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}
+        for i in s:
+            if i in numdict.keys():
+                num.append(numdict[i])
+            elif i == '+' or i == '-':
+                continue
+            else: 
+                return 0
+        ans = 0
+        if len(num) == 1 and num[0] == 0:
+            flag = True
+            return 0
+        for i in num:
+            ans = ans*10 + i
+        if s[0] == '-':
+            ans = 0 - ans
+        return ans

docs/剑指offer/Python/50-数组中重复的数字.py

@@ -0,0 +1,20 @@
+# -*- coding:utf-8 -*-
+class Solution:
+    # 这里要特别注意~找到任意重复的一个值并赋值到duplication[0]
+    # 函数返回True/False
+    def duplicate(self, numbers, duplication):
+        # write code here
+        if not numbers or len(numbers) < 0:
+            return False
+        for i in numbers:
+            if i < 0 or i > len(numbers) - 1:
+                return False
+        for i in range(len(numbers)):
+            while numbers[i] != i:
+                if numbers[i] == numbers[numbers[i]]:
+                    duplication[0] = numbers[i]
+                    return True
+                else:
+                    idx = numbers[i]
+                    numbers[i],numbers[idx] = numbers[idx],numbers[i]
+        return False

docs/剑指offer/Python/51-构建乘积数组.py

@@ -0,0 +1,15 @@
+# -*- coding:utf-8 -*-
+class Solution:
+    def multiply(self, A):
+        # write code here
+        if not A or len(A) <= 0:
+            return
+        length = len(A)
+        lis = [1] * length
+        for i in range(1,length):
+            lis[i] = lis[i-1] * A[i-1]
+        temp = 1
+        for i in range(length-2,-1,-1):
+            temp = temp * A[i+1]
+            lis[i] *= temp
+        return lis

docs/剑指offer/Python/52-正则表达式匹配.py

@@ -0,0 +1,36 @@
+# -*- coding:utf-8 -*-
+class Solution:
+    # s, pattern都是字符串
+    def match(self, s, pattern):
+        # 如果s与pattern都为空，则True
+        if len(s) == 0 and len(pattern) == 0:
+            return True
+        # 如果s不为空，而pattern为空，则False
+        elif len(s) != 0 and len(pattern) == 0:
+            return False
+        # 如果s为空，而pattern不为空，则需要判断
+        elif len(s) == 0 and len(pattern) != 0:
+            # pattern中的第二个字符为*，则pattern后移两位继续比较
+            if len(pattern) > 1 and pattern[1] == '*':
+                return self.match(s, pattern[2:])
+            else:
+                return False
+        # s与pattern都不为空的情况
+        else:
+            # pattern的第二个字符为*的情况
+            if len(pattern) > 1 and pattern[1] == '*':
+                # s与pattern的第一个元素不同，则s不变，pattern后移两位，相当于pattern前两位当成空
+                if s[0] != pattern[0] and pattern[0] != '.':
+                    return self.match(s, pattern[2:])
+                else:
+                    # 如果s[0]与pattern[0]相同，且pattern[1]为*，这个时候有三种情况
+                    # pattern后移2个，s不变；相当于把pattern前两位当成空，匹配后面的
+                    # pattern后移2个，s后移1个；相当于pattern前两位与s[0]匹配
+                    # pattern不变，s后移1个；相当于pattern前两位，与s中的多位进行匹配，因为*可以匹配多位
+                    return self.match(s, pattern[2:]) or self.match(s[1:], pattern[2:]) or self.match(s[1:], pattern)
+            # pattern第二个字符不为*的情况
+            else:
+                if s[0] == pattern[0] or pattern[0] == '.':
+                    return self.match(s[1:], pattern[1:])
+                else:
+                    return False