LeetCode（剑指 Offer）- 29. 顺时针打印矩阵

Posted 2022-04-15 放羊的牧码

题目链接：点击打开链接

题目大意：略。

解题思路：略。

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AC 代码

• Java

// 解决方案(1)
class Solution 
    public int[] spiralOrder(int[][] matrix) 
        if(matrix.length == 0) return new int[0];
        int l = 0, r = matrix[0].length - 1, t = 0, b = matrix.length - 1, x = 0;
        int[] res = new int[(r + 1) * (b + 1)];
        while(true) 
            for(int i = l; i <= r; i++) res[x++] = matrix[t][i]; // left to right
            if(++t > b) break;
            for(int i = t; i <= b; i++) res[x++] = matrix[i][r]; // top to bottom
            if(l > --r) break;
            for(int i = r; i >= l; i--) res[x++] = matrix[b][i]; // right to left
            if(t > --b) break;
            for(int i = b; i >= t; i--) res[x++] = matrix[i][l]; // bottom to top
            if(++l > r) break;
        
        return res;
    


// 解决方案(2)
class Solution 

    public int[] spiralOrder(int[][] matrix) 
        if (matrix.length == 0) return new int[0];
        int m = matrix.length, n = matrix[0].length;
        final int[] direct = 6, 2, 4, 8;
        int[][] block = new int[m][n];
        int[] res = new int[m * n];
        int i = 0, j = 0, q = 0, p = 0, dir;
        while (true) 
            dir = direct[p];
            if (dir == 6)  // RIGHT
                if (j >= n || block[i][j] == 1) 
                    p = (p + 1) % 4;
                    j--;
                    i++;
                    continue;
                
                block[i][j] = 1;
                res[q++] = matrix[i][j++];
             else if (dir == 2)  // DOWN
                if (i >= m || block[i][j] == 1) 
                    p = (p + 1) % 4;
                    i--;
                    j--;
                    continue;
                
                block[i][j] = 1;
                res[q++] = matrix[i++][j];
             else if (dir == 4)  // LEFT
                if (j < 0 || block[i][j] == 1) 
                    p = (p + 1) % 4;
                    j++;
                    i--;
                    continue;
                
                block[i][j] = 1;
                res[q++] = matrix[i][j--];
             else  // UP
                if (i < 0 || block[i][j] == 1) 
                    p = (p + 1) % 4;
                    i++;
                    j++;
                    continue;
                
                block[i][j] = 1;
                res[q++] = matrix[i--][j];
            
            if (q == m * n) 
                break;
            
        
        return res;
    

• C++

class Solution 
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix)
    
        if (matrix.empty()) return ;
        int l = 0, r = matrix[0].size() - 1, t = 0, b = matrix.size() - 1;
        vector<int> res;
        while(true)
        
            for (int i = l; i <= r; i++) res.push_back(matrix[t][i]); // left to right
            if (++t > b) break;
            for (int i = t; i <= b; i++) res.push_back(matrix[i][r]); // top to bottom
            if (l > --r) break;
            for (int i = r; i >= l; i--) res.push_back(matrix[b][i]); // right to left
            if (t > --b) break;
            for (int i = b; i >= t; i--) res.push_back(matrix[i][l]); // bottom to top
            if (++l > r) break;
        
        return res;
    
;