力扣题目解答自我总结（反转类题目）

一.反转字符串

1.题目描述

编写一个函数，其作用是将输入的字符串反转过来。输入字符串以字符数组 char[] 的形式给出。

不要给另外的数组分配额外的空间，你必须原地修改输入数组、使用 o(1) 的额外空间解决这一问题。

你可以假设数组中的所有字符都是 ascii 码表中的可打印字符。

示例 1：

输入：["h","e","l","l","o"]
输出：["o","l","l","e","h"]

示例 2：

输入：["h","a","n","n","a","h"]
输出：["h","a","n","n","a","h"]

2.解答

class solution:
    def reversestring(self, s: list[str]) -> none:
        """
        do not return anything, modify s in-place instead.
        """
        st_num = 0
        e_num = len(s)-1
        while e_num >st_num:
            s[st_num],s[e_num] = s[e_num],s[st_num]
            st_num += 1
            e_num -= 1
#难点就是在o(1) 下运行

二.反转整数

1.题目描述

给出一个 32 位的有符号整数，你需要将这个整数中每位上的数字进行反转。

示例 1:

输入: 123
输出: 321

示例 2:

输入: -123
输出: -321

示例 3:

输入: 120
输出: 21

2.解答

class solution:
    def reverse(self, x: int) -> int:
        new_x = '-'
        x = str(x)
        x = x[::-1]
        while x[-1] == 0:  #删除最后一位的 0
            x = x[:-1]
        if x[-1] == '-':   #删除有括号的
            x = x[:-1]
            while x[-1] == 0:
                x = x[:-1]
            new_x += x
            x = new_x
        if int(x) < -2**31 or int(x) >2**31-1:
            return 0
        else:
            return int(x)

三.旋转图像

1.题目描述

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]
]

2.解答

class solution:
    def rotate(self, matrix: list[list[int]]) -> none:
        """
        do not return anything, modify matrix in-place instead.
        """
        import copy
        new_list = []
        time = 0
        conter = 0
        matrix.reverse()
        matrix_1 = copy.copy(matrix)
        print(matrix)
        if len(matrix) != 0:
            for a in range(len(matrix)):
                matrix[a] = []
            while len(matrix) > len(matrix_1[0]):
                matrix.pop()
            while len (matrix) < len(matrix_1[0]):
                matrix.append([])
            for b in range(len(matrix_1[0])):
                for c in range(len(matrix_1)):
                    if time != len(matrix_1):
                        time += 1
                        matrix[conter].append(matrix_1[c][b])
                    elif time == len(matrix_1):
                        time = 1
                        conter += 1
                        matrix[conter].append(matrix_1[c][b])
#这是我的思路比较low，先水平翻转，再按照子列表的长度，把他变成有拥有子列表长度一样的空的列表，再里面填写值进去，写完了头有点昏，等下次有空再优化下代码

四.反转字符串中的单词 ⅲ

1.题目描述

给定一个字符串，你需要反转字符串中每个单词的字符顺序，同时仍保留空格和单词的初始顺序。

示例 1:

输入: "let's take leetcode contest"
输出: "s'tel ekat edocteel tsetnoc" 

注意：在字符串中，每个单词由单个空格分隔，并且字符串中不会有任何额外的空格。

2.解答

class solution:
    def reversewords(self, s: str) -> str:
        s_1 = ''
        for i in s.split():
            s_1 += i[::-1]+' '
        s_1 = s_1[:-1]
        return s_1
    #这个写法比较low

五.有效数独

1.题目描述

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。

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 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。

2.解答

class solution:
    def isvalidsudoku(self, board: list[list[str]]) -> bool:
        import copy
        x = true
        s_1 = ''
        s_2 = ''
        new_list = copy.deepcopy(list)
        list_1 = copy.copy(list)
        for num_1 in range(9):
            for num_2 in range(9):
                new_list[num_1][num_2] = list[num_2][num_1]
        new_list_1 =copy.copy(new_list)
        for a_1 in range(9):
            for a_2 in range(9):
                s_1 += str(list[a_1][a_2])
                list_1[a_1] = s_1
                s_2 += str(new_list[a_1][a_2])
                new_list_1[a_1] = s_2
                if len(list_1[a_1]) == 9:
                    list_1[a_1] = list_1[a_1].replace('.','')
                    new_list_1[a_1] = new_list_1[a_1].replace('.', '')
                    s_1 = ''
                    s_2 = ''
                    num_1 = len(list_1[a_1])-len(set(list_1[a_1]))
                    num_2 = len(new_list_1[a_1])-len(set(new_list_1[a_1]))
                    if num_1 != 0 or num_2 != 0:
                        x = false
        return x
    
    #pycharm能运行，力扣里就不行，很闷逼大神路过留个言
    #下面是力扣里报错内容
    #line 11: typeerror: parameters to generic types must be types. got 0.

如果你有更加吊炸天的解题方法留言，让我这渣渣学学```