二刷 leetcode（4）

36. 有效的数独

判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。

数字 1-9 在每一行只能出现一次。数字 1-9 在每一列只能出现一次。数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字，空白格用 '.' 表示。

示例 1:

输入: [ ["5","3",".",".","7",".",".",".","."], ["6",".",".","1","9","5",".",".","."], [".","9","8",".",".",".",".","6","."], ["8",".",".",".","6",".",".",".","3"], ["4",".",".","8",".","3",".",".","1"], ["7",".",".",".","2",".",".",".","6"], [".","6",".",".",".",".","2","8","."], [".",".",".","4","1","9",".",".","5"], [".",".",".",".","8",".",".","7","9"] ] 输出: true

示例 2:

输入: [   ["8","3",".",".","7",".",".",".","."],   ["6",".",".","1","9","5",".",".","."],   [".","9","8",".",".",".",".","6","."],   ["8",".",".",".","6",".",".",".","3"],   ["4",".",".","8",".","3",".",".","1"],   ["7",".",".",".","2",".",".",".","6"],   [".","6",".",".",".",".","2","8","."],   [".",".",".","4","1","9",".",".","5"],   [".",".",".",".","8",".",".","7","9"] ] 输出: false 解释: 除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

说明:

一个有效的数独（部分已被填充）不一定是可解的。只需要根据以上规则，验证已经填入的数字是否有效即可。给定数独序列只包含数字 1-9 和字符 '.' 。给定数独永远是 9x9 形式的。

代码思路：

先用一个函数判断是不是有效行(列)，同时判断这一行(列)有没有重复的元素。

class Solution: def isValidSudoku(self, board): """ :type board: List[List[str]] :rtype: bool """ def isValidList(xs): xs = list(filter(lambda x: x !="." ,xs)) return len(set(xs)) == len(xs) for i in range(9): if not isValidList([board[i][j] for j in range(9)]) or not isValidList([board[j][i] for j in range(9)]): return False for i in range(3): for j in range(3): if not isValidList([board[m][n] for n in range(3 * j, 3 * j + 3) for m in range(3 * i, 3*i+3)]): return False return True

367. 有效的完全平方数

给定一个正整数 num，编写一个函数，如果 num 是一个完全平方数，则返回 True，否则返回 False。

注意：不要使用任何内置的库函数，如  sqrt。

示例 1：

输入： 16 输出： True

 示例 2：

输入： 14 输出： False

代码：前两种方法超时了，第三种方法改进

class Solution: def isPerfectSquare(self, num): """ :type num: int :rtype: bool """ #### 超出时间限制 # if num == 1: # return True # m = num //2 # for i in range(m+1): # if i**2 == num: # return True # return False #### 通过列举所有的完全平方数，1，4，9，16，25，36，49，64，81，100…等等， #### 发现完全平方数的差都为奇数，即1，3，5，7，9，11，13，15… #### 等等~所以可以判断完全平方数应该是N个奇数的和。 #### 超出时间限制 # if num == 1: # return True # for i in range(1, num, 2): # num -= i # if num == 0: # return True # else: # continue # return False left, right = 0, num while left < right: mid = left + (right - left) // 2 if mid ** 2 < num: left = mid + 1 else: right = mid if left < 1: sqrt_num = left -1 else: sqrt_num = left return sqrt_num ** 2 == num

48. 旋转图像

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给定一个 n × n 的二维矩阵表示一个图像。

将图像顺时针旋转 90 度。

说明：

你必须在原地旋转图像，这意味着你需要直接修改输入的二维矩阵。请不要使用另一个矩阵来旋转图像。

示例 1:

给定 matrix = [ [1,2,3], [4,5,6], [7,8,9] ], 原地旋转输入矩阵，使其变为: [ [7,4,1], [8,5,2], [9,6,3] ]

示例 2:

给定 matrix = [ [ 5, 1, 9,11], [ 2, 4, 8,10], [13, 3, 6, 7], [15,14,12,16] ], 原地旋转输入矩阵，使其变为: [ [15,13, 2, 5], [14, 3, 4, 1], [12, 6, 8, 9], [16, 7,10,11] ]

代码：

class Solution: def rotate(self, matrix): """ :type matrix: List[List[int]] :rtype: void Do not return anything, modify matrix in-place instead. """ ##### the first method # n = len(matrix) # m = n // 2 # for i in range(n): # for j in range(n-i-1): # temp = matrix[i][j] # matrix[i][j] = matrix[n-j-1][n-i-1] # matrix[n-j-1][n-i-1] = temp # for i in range(m): # for j in range(n): # temp = matrix[i][j] # matrix[i][j] = matrix[n-i-1][j] # matrix[n-i-1][j] = temp ##### the second method n = len(matrix) for i in range(n): for j in range(i+1, n): matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j] print(matrix) for i in range(n): matrix[i].reverse() print(matrix)