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

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)

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