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。
- 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。
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```