leetcode题库解答源码(python3)
下面和大家分享本人在leetcode上已经ace的题目源码(python3): 本人会持续更新!~
class Leetcode_Solution(object):
def twoSum_1(self,nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
'''
# 此解法复杂度为O(n^2)
new_nums = []
for i in range(len(nums)):
for j in range(i+1,len(nums)):
if nums[i] + nums[j] == target:
new_nums.append(i)
new_nums.append(j)
return new_nums
'''
# 此解法复杂度为O(n)
# 拓展:若解不唯一,可先将nums排序后进行下面操作,将全部符合对输出
if len(nums)<= 1:
return False
else:
dict = {}
for i in range(len(nums)):
# 字典底层是用hash表实现的,无论字典中有多少元素,查找的平云复杂度均为O(1)
if num[i] in dict:
return [dict[nums[i]], i]
else:
dict[target - nums[i]] = i
def reverse_7(self,x):
"""
:type x: int
:rtype: int
"""
MAX = 2**31 - 1
min = -1*2**31
if x < 0:
y = -1*int(str(-x)[::-1])
else:
y = int(str(x)[::-1])
if y > Max or y < min:
return 0
return y
def isPalindrome_9(self, x):
renum = 0
if x < 0 or (x % 10 == 0 and x != 0):
return False
while x > renum:
renum = renum * 10 + x % 10
x /= 10
return x == renum or x == renum/10
def romanToInt_13(self, s):
"""
:type s: str
:rtype: int
"""
dic = {'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}
sum = 0
for i in range(len(s)-1):
if dic[s[i]] < dic[s[i+1]]:
sum -= dic[s[i]]
else:
sum += dic[s[i]]
return sum + dic[s[-1]]
def longestCommonPrefix_14(self, strs):
"""
:type strs: List[str]
:rtype: str
"""
if len(strs) == 0: # Horizontal scanning/////another way: vertical scanning
return ''
prefix = strs[0]
for i in range(1,len(strs)):
while strs[i].find(prefix) != 0:
prefix = prefix[0:len(prefix)-1]
if prefix == '':
return ''
return prefix
def isValid_20(self, s):
"""
:type s: str
:rtype: bool
"""
'''
list = []
a = b = c = 0
if len(s) == 0:
return True
for i in range(len(s)):
if s[i] == '(':
list.append(s[i])
a += 1
if s[i] == '{':
list.append(s[i])
b += 1
if s[i] == '[':
list.append(s[i])
c += 1
if s[i] == ')':
if len(list) != 0 and list[-1] == '(':
list.pop()
a -= 1
else:
return False
if s[i] == '}':
if len(list) != 0 and list[-1] == '{':
list.pop()
b -= 1
else:
return False
if s[i] == ']':
if len(list) != 0 and list[-1] == '[':
list.pop()
c -= 1
else:
return False
if len(list) == 0 and a == b == c == 0:
return True
else:
return False
'''
dic = {')':'(','{':'}','[':']'}
stack = []
for i in s:
if i in dic.values():
stack.append(i)
elif i in dic.keys():
if stack == [] or dic[i] != stack.pop():
return False
else:
return False
return stack == []
def mergeTwoLists_21(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
head = rear = ListNode(0)
while l1 and l2:
if l1.val < l2.val:
rear.next = l1
l1 = l1.next
else:
rear.next = l2
l2 = l2.next
rear = rear.next
rear.next = l1 or l2
return head.next
def removeDuplicates_26(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) == 0:
return 0
newtail = 0
for i in range(1,len(nums)):
if nums[i] != nums[newtail]:
newtail += 1
nums[newtail] = nums[i]
return newtail + 1
def removeElement_27(self, nums, val):
"""
:type nums: List[int]
:type val: int
:rtype: int
"""
i = len(nums)
j = 0
if i == 0:
return 0
while j < i:
if nums[j] == val:
nums.pop(j)
i -= 1
else:
j += 1
return len(nums)
def strStr_28(self, haystack, needle):
"""
:type haystack: str
:type needle: str
:rtype: int
"""
for i in range(len(haystack) - len(needle) +1):
if haystack[i:i+len(needle)] == needle:
return i
return -1
def searchInsert_35(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
return len([x for x in nums if x < target])
def countAndSay(self, n):
"""
:type n: int
:rtype: str
"""
------口天丶木乔