这篇文章包含三个编程问题的详细描述：冒泡排序实现，检查9x9数独的有效性，以及在有序数组中删除重复元素（最多保留两次）。每个问题都提供了输入输出示例和部分代码实现，旨在作为每日一练的编程题目供学习者练习。
目录
2. 有效的数独 🌟🌟
1. 冒泡排序法排序
输入n（1≤n≤10）个整数，用冒泡排序法对其从小到大排序，共进行n-1趟，要求输出每一趟的排序情况
输入格式:
先输入个数n，再输入n个整数。
输出格式:
第1趟结果
第2趟结果
......
第n-1趟结果
每个数后面有一个空格，每个序列占一行。
输入样例:
5
4 2 3 2 1输出样例:
2 3 2 1 4
2 2 1 3 4
2 1 2 3 4
1 2 2 3 4代码：
#include "stdio.h"
int main()
{
int arr[10];
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%d", &arr[i]);
for (int i = 0; i < n - 1; i++)
{
for (int j = 0; j < n - i - 1; j++)
{
if (arr[j] > arr[j + 1])
{
int t = arr[j];
arr[j] = arr[j + 1];
arr[j + 1] = t;
}
}
for (int j = 0; j < n; j++)
printf("%d ", arr[j]);
printf("\n");
}
return 0;
}输入输出：
5
4 2 3 2 1
2 3 2 1 4
2 2 1 3 4
2 1 2 3 4
1 2 2 3 4
--------------------------------
Process exited after 12.95 seconds with return value 0
请按任意键继续. . .
2. 有效的数独
请你判断一个 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。（请参考示例图）
数独部分空格内已填入了数字，空白格用 '.' 表示。
注意：
- 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。
示例 1：
输入：board = [["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：board = [["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。
提示：
board.length == 9board[i].length == 9board[i][j]是一位数字或者'.'
代码：
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
bool isValidSudoku(vector<vector<char>> &board)
{
for (int i = 0; i < board.size(); i++)
{
vector<bool> mark(10);
for (int j = 0; j < board.size(); j++)
{
if (!valid(board, mark, i, j))
{
return false;
}
}
}
for (int j = 0; j < board.size(); j++)
{
vector<bool> mark(10);
for (int i = 0; i < board.size(); i++)
{
if (!valid(board, mark, i, j))
{
return false;
}
}
}
for (int k = 0; k < board.size(); k++)
{
int sr = k / 3 * 3;
int sc = (k % 3) * 3;
vector<bool> mark(10);
for (int i = sr; i < sr + 3; i++)
{
for (int j = sc; j < sc + 3; j++)
{
if (!valid(board, mark, i, j))
{
return false;
}
}
}
}
return true;
}
private:
bool valid(vector<vector<char>> &board, vector<bool> &mark, int i, int j)
{
if (board[i][j] != '.')
{
int index = board[i][j] - '0';
if (mark[index])
{
return false;
}
else
{
mark[index] = 1;
}
}
return true;
}
};
int main()
{
Solution s;
vector<vector<char>> board = {
{'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'}};
cout << s.isValidSudoku(board) << endl;
board = {
{'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'}};
cout << s.isValidSudoku(board) << endl;
return 0;
} 输出：
1
0
3. 删除有序数组中的重复项 II
给你一个有序数组 nums ，请你 原地 删除重复出现的元素，使每个元素 最多出现两次 ，返回删除后数组的新长度。
不要使用额外的数组空间，你必须在 原地 修改输入数组 并在使用 O(1) 额外空间的条件下完成。
说明：
为什么返回数值是整数，但输出的答案是数组呢？
请注意，输入数组是以「引用」方式传递的，这意味着在函数里修改输入数组对于调用者是可见的。
你可以想象内部操作如下:
// nums 是以“引用”方式传递的。也就是说，不对实参做任何拷贝
int len = removeDuplicates(nums);// 在函数里修改输入数组对于调用者是可见的。
// 根据你的函数返回的长度, 它会打印出数组中 该长度范围内 的所有元素。
for (int i = 0; i < len; i++) {
print(nums[i]);
}
示例 1：
输入：nums = [1,1,1,2,2,3] 输出：5, nums = [1,1,2,2,3] 解释：函数应返回新长度 length = 5, 并且原数组的前五个元素被修改为 1, 1, 2, 2, 3 。 不需要考虑数组中超出新长度后面的元素。
示例 2：
输入：nums = [0,0,1,1,1,1,2,3,3] 输出：7, nums = [0,0,1,1,2,3,3] 解释：函数应返回新长度 length = 7, 并且原数组的前五个元素被修改为 0, 0, 1, 1, 2, 3, 3 。 不需要考虑数组中超出新长度后面的元素。
提示：
1 <= nums.length <= 3 * 10^4-10^4 <= nums[i] <= 10^4nums已按升序排列
#include <stdio.h>
#include <stdlib.h>
static int removeDuplicates(int *nums, int numsSize)
{
if (numsSize == 0)
{
return 0;
}
int i;
int len = 0;
int count = 1;
for (i = 1; i < numsSize; i++)
{
if (nums[len] == nums[i])
{
if (count < 2)
{
count++;
nums[++len] = nums[i];
}
}
else
{
count = 1;
nums[++len] = nums[i];
}
}
return len + 1;
}
void printResult(int *nums, int count)
{
printf("%d, nums = [", count);
for (int i = 0; i < count; i++)
{
printf("%d", nums[i]);
if (i < count - 1)
printf(",");
}
printf("]\n");
}
int main()
{
int count = 6;
int nums[count] = {1,1,1,2,2,3};
count = removeDuplicates(nums, count);
printResult(nums, count);
count = 9;
int nums2[count] = {0,0,1,1,1,1,2,3,3};
count = removeDuplicates(nums2, count);
printResult(nums2, count);
return 0;
}输出：
5, nums = [1,1,2,2,3]
7, nums = [0,0,1,1,2,3,3]
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