leetcode practice - python3 (8)
169. Majority Element
Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
You may assume that the array is non-empty and the majority element always exist in the array.
Example 1:
Input: [3,2,3]
Output: 3
Example 2:
Input: [2,2,1,1,1,2,2]
Output: 2
思路：编程之美发帖水王
class Solution:
def majorityElement(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) <= 2:
return nums[0]
cur, cnt = nums[0], 1
for n in nums[1 : ]:
if n != cur:
cnt -= 1
if cnt == 0:
cur, cnt = n, 1
else:
cnt += 1
return curBeat 13.69% python3 2018-05-25
class Solution:
def majorityElement(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
cnt, cur = 0, None
for n in nums:
if cnt == 0:
cur = n
if n == cur:
cnt += 1
else:
cnt -= 1
return curBeat 91.20% python3 2018-05-25
class Solution:
def majorityElement(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
return sorted(nums)[len(nums) // 2]Beat 98.76% python3 2018-05-25
好吧，排序反而更快…
206. Reverse Linked List
Reverse a singly linked list.
Example:
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL
Follow up:
A linked list can be reversed either iteratively or recursively. Could you implement both?
思路：断开链表，保存新head和老头部，每次把老头部的链到新头部前面
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if head == None or head.next == None:
return head
p, q = head, head.next
head.next = None
while q:
p = q
q = q.next
p.next = head
head = p
return headBeat 99.86% python3 2018-05-25
注：多次提交时间不同
226. Invert Binary Tree
Invert a binary tree.
Example:
Input:
4
/ \
2 7
/ \ / \
1 3 6 9
Output:
4
/ \
7 2
/ \ / \
9 6 3 1
Trivia:
This problem was inspired by this original tweet by Max Howell:
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.
思路：DFS
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def invertTree(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if root is None:
return root
self.invertTree(root.left)
self.invertTree(root.right)
root.left, root.right = root.right, root.left
return rootBeat 99.43% python3 2018-05-25
234. Palindrome Linked List
Given a singly linked list, determine if it is a palindrome.
Example 1:
Input: 1->2
Output: false
Example 2:
Input: 1->2->2->1
Output: true
Follow up:
Could you do it in O(n) time and O(1) space?
思路：找到中点，后半部分reverse，比较
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def revertList(self, head):
if head is None or head.next is None:
return head
p, q = head, head.next
head.next = None
while q:
p = q
q = q.next
p.next = head
head = p
return head
def isPalindrome(self, head):
"""
:type head: ListNode
:rtype: bool
"""
if head == None or head.next == None:
return True
slow, fast = head, head
while fast.next and fast.next.next:
slow, fast = slow.next, fast.next.next
revert = self.revertList(slow.next)
while revert:
if revert.val != head.val:
return False
head, revert = head.next, revert.next
return TrueBeat 96.70% python3 2018-05-25
283. Move Zeroes
Given an array nums, write a function to move all 0’s to the end of it while maintaining the relative order of the non-zero elements.
Example:
Input: [0,1,0,3,12]
Output: [1,3,12,0,0]
Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.
简单思路：两个指针，一个指向0，一个指向非0，交换
class Solution:
def moveZeroes(self, nums):
"""
:type nums: List[int]
:rtype: void Do not return anything, modify nums in-place instead.
"""
length = len(nums)
if length <= 1:
return
left, right = 0, 0
while left < length and right < length:
while left < length and nums[left] != 0:
left += 1
if right < left:
right = left
while right < length and nums[right] == 0:
right += 1
if left < right and right < length:
nums[left] = nums[right]
nums[right] = 0
returnBeat 33.86% python3 2018-05-25
优化：更简单的写法
class Solution:
def moveZeroes(self, nums):
"""
:type nums: List[int]
:rtype: void Do not return anything, modify nums in-place instead.
"""
pos = 0
for i in range(len(nums)):
if nums[i] != 0:
nums[pos], nums[i] = nums[i], nums[pos]
pos += 1Beat 99.30% python3 2018-05-25