有效的数独 LeetCode(数组的巧妙运用)

本文介绍了一种有效的数独验证算法,通过使用map和三维数组优化检查过程,确保数独中1-9数字在每行、每列及每个3x3宫内不重复。文章对比了两种实现方式,并分析了其运行效率。

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  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。

  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。

题解:

今天刷了好几道题,就这题还有点意思,主要是优化,横纵坐标的有效性没问题直接用map搞,3x3的方格的有效性我一开始是直接写了个check函数,在函数里用map进行验证,后来看见别人的代码,觉得很巧妙,可以开一个三维数组,假设验证第i行第j个,那么可以把它唯一标记为p[i/3][j/3],可以验证(0,0),(0,1)一直到(2,2)也就是第一个方格所有坐标都符合这个唯一标记,而其他的坐标不符合,那么开一个三维数组就可以标记整张图的每个3x3的方格里的有效性。我修改之后的时间居然跑得和原来的一样。。莫名其妙

我修改前的代码:

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        for(int i=0;i<9;i++)
        {
            map<char,int>p;
            map<char,int>p1;
            for(int j=0;j<9;j++)
            {
                if(i%3==0&&j%3==0)
                {
                    if(!check(board,i,j))
                        return false;
                }
                if(board[i][j]!='.')
                {
                    if(p[board[i][j]])
                        return false;
                    p[board[i][j]]=1;
                }
                if(board[j][i]!='.')
                {
                    if(p1[board[j][i]])
                        return false;
                    p1[board[j][i]]=1;
                }
            }
        }
        return true;
    }
    bool check(vector<vector<char>>& board,int sx,int sy)
    {
        map<char,int>p;
        for(int i=sx;i<sx+3;i++)
        {
            for(int j=sy;j<sy+3;j++)
            {
                if(board[i][j]=='.')
                    continue;
                if(p[board[i][j]])
                    return false;
                p[board[i][j]]=1;
            }
        }
        return true;
    }
};

我修改后的代码:

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int p2[10][10][10];
        memset(p2,0,sizeof(p2));
        for(int i=0;i<9;i++)
        {
            map<char,int>p;
            map<char,int>p1;
            for(int j=0;j<9;j++)
            {
                if(board[i][j]!='.')
                {
                    if(p[board[i][j]])
                        return false;
                    p[board[i][j]]=1;
                    int t=board[i][j]-'0';
                    if(p2[i/3][j/3][t])
                        return false;
                    p2[i/3][j/3][t]=1;
                }
                if(board[j][i]!='.')
                {
                    if(p1[board[j][i]])
                        return false;
                    p1[board[j][i]]=1;
                }
            }
        }
        return true;
    }
};

 

### LeetCode 数组题目及解 #### 二分查找 在处理有序数组时,二分查找是一种高效的算。通过不断将搜索范围减半来快速定位目标值的位置[^1]。 ```cpp int binarySearch(int arr[], int l, int r, int x) { while (l <= r) { int m = l + (r - l) / 2; // Check if x is present at mid if (arr[m] == x) return m; // If x greater, ignore left half if (arr[m] < x) l = m + 1; // If x is smaller, ignore right half else r = m - 1; } // if we reach here, then element was not present return -1; } ``` #### 移除元素 针对移除指定元素的任务,可以采用双指针方提高效率。这种方不仅减少了不必要的遍历次数,还简化了代码逻辑[^3]。 ```cpp class Solution { public: int removeElement(vector<int>& nums, int val) { int k = 0; for (int i = 0; i < nums.size(); ++i) { if (nums[i] != val) { nums[k++] = nums[i]; } } return k; } }; ``` #### 查找重复数字 当面对只读且不允许额外空间开销的情况下寻找重复项的问题时,可以通过巧妙利用原数组特性实现线性时间内解决问题的方[^2]。 ```cpp class Solution { public: int findDuplicate(const vector<int>& nums) const { int slow = nums[0], fast = nums[nums[0]]; while(slow != fast){ slow = nums[slow]; fast = nums[nums[fast]]; } fast = 0; while(fast != slow){ slow = nums[slow]; fast = nums[fast]; } return slow; } }; ``` 这些例子展示了如何运用不同策略解决LeetCode上的典型数组类问题。每种解决方案都考虑到了特定条件下的最优实践,旨在帮助开发者提升编程技巧并加深对数据结构的理解。
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