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Commit fa61f11

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nature1995nature1995
committed
Update 003
Update 003
1 parent 28ddde6 commit fa61f11
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4 files changed

+106
-11
lines changed

‎Python3/001_Two_Sum.py

Copy file name to clipboardExpand all lines: Python3/001_Two_Sum.py
+2Lines changed: 2 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -1,3 +1,5 @@
1+
#!usr/bin/env python
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# -*- coding:utf-8 -*-
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'''
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Given an array of integers, find two numbers such that they add up to a specific target number.
35

‎Python3/002_Add_Two_Numbers.py

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+15-11Lines changed: 15 additions & 11 deletions
Original file line numberDiff line numberDiff line change
@@ -1,3 +1,5 @@
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#!usr/bin/env python
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# -*- coding:utf-8 -*-
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'''
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You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
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@@ -11,11 +13,10 @@ def __init__(self, x):
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self.val = x
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self.next = None
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# Define this to check if it works well
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def myPrint(self):
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def my_test(self):
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print(self.val)
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if self.next:
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self.next.myPrint()
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self.next.my_test()
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class Solution(object):
@@ -25,15 +26,14 @@ def addTwoNumbers(self, l1, l2):
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:type l2: ListNode
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:rtype: ListNode
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"""
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result = ListNode(0);
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cur = result;
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result = ListNode(0)
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cur = result
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while l1 or l2:
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cur.val += self.addTwoNodes(l1, l2)
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cur.val += self.AddTwoNodes(l1, l2)
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if cur.val >= 10:
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cur.val -= 10
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cur.next = ListNode(1)
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else:
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# Check if there is need to make the next node
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if l1 and l1.next or l2 and l2.next:
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cur.next = ListNode(0)
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cur = cur.next
@@ -43,9 +43,9 @@ def addTwoNumbers(self, l1, l2):
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l2 = l2.next
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return result
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def addTwoNodes(self, n1, n2):
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def AddTwoNodes(self, n1, n2):
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if not n1 and not n2:
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# This cannot happen, ignore it
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None
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if not n1:
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return n2.val
@@ -54,12 +54,16 @@ def addTwoNodes(self, n1, n2):
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return n1.val + n2.val
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if __name__ == "__main__":
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list = ListNode(2)
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list.next = ListNode(4)
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list.next.next = ListNode(3)
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list1= ListNode(5)
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list1 = ListNode(5)
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list1.next = ListNode(6)
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list1.next.next = ListNode(4)
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65-
print(Solution().addTwoNumbers(list, list1).myPrint())
69+
print(Solution().addTwoNumbers(list, list1).my_test())

‎Python3/003_Longest_Substring_Without_Repeating_Characters.py

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+35Lines changed: 35 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,35 @@
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#!usr/bin/env python
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# -*- coding:utf-8 -*-
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'''
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Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.
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'''
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class Solution(object):
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def lengthOfLongestSubstring(self, s):
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"""
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:type s: str
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:rtype: int
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"""
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if not s:
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return 0
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if len(s) <= 1:
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return len(s)
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locations = [-1 for i in range(256)]
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print(locations)
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index = -1
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m = 0
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for i, v in enumerate(s):
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#如果出现的字符就需要更新index为当前字符的位置
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if (locations[ord(v)] > index):
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index = locations[ord(v)]
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m = max(m, i - index)
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#确定字符的位置，相同字符的位置一样
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locations[ord(v)] = i
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return m
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32+
if __name__ == "__main__":
33+
print(Solution().lengthOfLongestSubstring("bccab"))
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‎Python3/004_Median_of_Two_Sorted_Arrays.py

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+54Lines changed: 54 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,54 @@
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#!usr/bin/env python
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# -*- coding:utf-8 -*-
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'''
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There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
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'''
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class Solution(object):
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def findMedianSortedArrays(self, nums1, nums2):
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"""
12+
:type nums1: List[int]
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:type nums2: List[int]
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:rtype: float
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"""
16+
length1 = len(nums1)
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length2 = len(nums2)
18+
k = (length1 + length2) // 2
19+
if (length1 + length2) % 2 == 0:
20+
return (self.findK(nums1, nums2, k) + self.findK(nums1, nums2, k - 1)) / 2.0; # 2 is enough in python3
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else:
22+
return self.findK(nums1, nums2, k)
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24+
def findK(self, num1, num2, k):
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# Recursive ends here
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if not num1:
27+
return num2[k]
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if not num2:
29+
return num1[k]
30+
if k == 0:
31+
return min(num1[0], num2[0])
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33+
length1 = len(num1)
34+
length2 = len(num2)
35+
if num1[length1 // 2] > num2[length2 // 2]:
36+
if k > length1 // 2 + length2 // 2:
37+
return self.findK(num1, num2[length2 // 2 + 1:], k - length2 // 2 - 1)
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else:
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return self.findK(num1[:length1 // 2], num2, k)
40+
else:
41+
if k > length1 // 2 + length2 // 2:
42+
return self.findK(num1[length1 // 2 + 1:], num2, k - length1 // 2 - 1)
43+
else:
44+
return self.findK(num1, num2[:length2 // 2], k)
45+
46+
47+
if __name__ == "__main__":
48+
assert Solution().findMedianSortedArrays([1, 2], [1, 2, 3]) == 2
49+
assert Solution().findMedianSortedArrays([], [2, 3]) == 2.5
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