力扣（2024.06.22）

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1. 59——螺旋矩阵2

给你一个正整数 n ，生成一个包含 1 到 n2 所有元素，且元素按顺时针顺序螺旋排列的 n x n 正方形矩阵 matrix 。

标签：数组，矩阵，模拟

代码：

class Solution:def generateMatrix(self, n: int) -> List[List[int]]:matrix = []for i in range(n):matrix.append([0] * n)top = 0bottom = n - 1left = 0right = n - 1num = 1while num <= n*n:for i in range(left, right + 1):matrix[top][i] = numnum = num + 1top = top + 1for i in range(top, bottom + 1):matrix[i][right] = numnum = num + 1right = right - 1for i in range(right, left - 1, -1):matrix[bottom][i] = numnum = num + 1bottom = bottom - 1for i in range(bottom, top - 1, -1):matrix[i][left] = numnum = num + 1left = left + 1return matrix

2. 60——排列序列

给出集合 [1,2,3,...,n]，其所有元素共有 n! 种排列。按大小顺序列出所有排列情况，并一一标记，当 n = 3 时, 所有排列如下：

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

给定 n 和 k，返回第 k 个排列。

标签：递归，数学

代码：

class Solution:def getPermutation(self, n: int, k: int) -> str:def fun(nums):n = len(nums) - 1while n > 0:if nums[n - 1] > nums[n]:n = n - 1else:breakidx1 = n - 1num1 = nums[idx1]n = len(nums) - 1while n > idx1:if nums[n] < num1:n = n - 1else:breakidx2 = nnum2 = nums[idx2]nums[idx1], nums[idx2] = num2, num1left = idx1 + 1right = len(nums) - 1while left < right:nums[left], nums[right] = nums[right], nums[left]left = left + 1right = right - 1return numsnums = []res = ""for i in range(n):nums.append(i + 1)i = 1while i < k:nums = fun(nums)i = i + 1for i in nums:res = res + str(i)return res

3. 61——旋转链表

给你一个链表的头节点 head ，旋转链表，将链表每个节点向右移动 k 个位置。

标签：链表，双指针

代码：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:pointer = headleng = 0while pointer:leng = leng + 1pointer = pointer.nextif leng == 0 or leng == 1:return headif k == 0 or k % leng == 0:return headk = k % leng  # 向右移动k次，对应倒数第k个结点会是新的头结点pointer = headwhile k:pointer = pointer.nextk = k - 1pointer1 = headwhile pointer.next:pointer1 = pointer1.nextpointer = pointer.nextres = pointer1.nextpointer1.next = Nonepointer.next = headreturn res

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