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 cur

Beat 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 cur

Beat 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:

扫描二维码关注公众号，回复： 1768384 查看本文章

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 head

Beat 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 root

Beat 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 True

Beat 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

        return

Beat 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 += 1

Beat 99.30% python3 2018-05-25