[LeetCode]剑指 Offer 29. 顺时针打印矩阵

Posted 2022-06-25 Spring-_-Bear

输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字。

示例 1：

输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[1,2,3,6,9,8,7,4,5]

示例 2：

输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
输出：[1,2,3,4,8,12,11,10,9,5,6,7]

限制：

0 <= matrix.length <= 100
0 <= matrix[i].length <= 100

题解：

/**
     * 剑指 Offer 29. 顺时针打印矩阵
     *
     * @param matrix 矩阵
     * @return 顺时针由外至内的矩阵遍历结果
     */
    public int[] spiralOrder(int[][] matrix) 
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) 
            return new int[0];
        
        int cols = matrix[0].length;
        int rows = matrix.length;
        int left = 0;
        int right = cols - 1;
        int up = 0;
        int down = rows - 1;
        /*      left         right
         * up   [1,  2,  3,  4],
         *      [5,  6,  7,  8],
         * down [9,  10, 11, 12]
         *
         */
        int[] res = new int[cols * rows];
        int index = 0;
        while (left <= right && up <= down) 
            // 从左向右
            for (int i = left; i <= right; i++, index++) 
                res[index] = matrix[up][i];
            
            // 从上至下
            for (int i = up + 1; i <= down; i++, index++) 
                res[index] = matrix[i][right];
            
            // 从右至左
            // if 判断处理只有一行或剩最后一行时，避免先 left->right，后 right->left 导致重复
            if (up != down) 
                for (int i = right - 1; i >= left; i--, index++) 
                    res[index] = matrix[down][i];
                
            
            // 从下至上
            // if 判断处理只有一列或剩最后一列时，避免先 up->down，后 down->up 导致重复
            if (left != right) 
                for (int i = down - 1; i > up; i--, index++) 
                    res[index] = matrix[i][left];
                
            
            left++;
            right--;
            up++;
            down--;
        
        return res;