Leetcode 1861. Rotating the Box

You are given an m x n matrix of characters box representing a side-view of a box. Each cell of the box is one of the following:

• A stone '#'
• A stationary obstacle '*'
• Empty '.'

The box is rotated 90 degrees clockwise, causing some of the stones to fall due to gravity. Each stone falls down until it lands on an obstacle, another stone, or the bottom of the box. Gravity does not affect the obstacles' positions, and the inertia from the box's rotation does not affect the stones' horizontal positions.

It is guaranteed that each stone in box rests on an obstacle, another stone, or the bottom of the box.

Return an n x m matrix representing the box after the rotation described above.

Example 1:

Input: box = [["#",".","#"]]
Output: [["."],
         ["#"],
         ["#"]]

Example 2:

Input: box = [["#",".","*","."],
              ["#","#","*","."]]
Output: [["#","."],
         ["#","#"],
         ["*","*"],
         [".","."]]

Example 3:

Input: box = [["#","#","*",".","*","."],
              ["#","#","#","*",".","."],
              ["#","#","#",".","#","."]]
Output: [[".","#","#"],
         [".","#","#"],
         ["#","#","*"],
         ["#","*","."],
         ["#",".","*"],
         ["#",".","."]]

Constraints:

• m == box.length
• n == box[i].length
• 1 <= m, n <= 500
• box[i][j] is either '#', '*', or '.'.

 【题目分析】

将一个矩阵顺时针旋转90度，矩阵中的石头在重力的牵引下向下掉落，直到遇到障碍物或者其他石块。

注意：

1. 矩阵是顺时针旋转，不是矩阵的转置。因此变换关系为，res[j][m-1-i] = box[i][j]，m = box.length.

【思路分析】

创建一个新矩阵，为新矩阵的每一列从下往上赋值，这个过程中有一个指针，一个记录经过变化后当前元素应该落在哪个位置，一个记录由于重力原因，该元素实际应该放置的位置。

【Java代码】

class Solution {
    public char[][] rotateTheBox(char[][] box) {
        int m = box.length, n = box[0].length;
        char[][] res = new char[n][m];
        for(int i = 0; i < m; i++) {
            for(int j = n-1, k = n-1; j >= 0; j--) {
                res[j][m-1-i] = '.';
                if(box[i][j] != '.') {
                    if(box[i][j] == '*') {
                        k = j;
                    }
                    res[k--][m-1-i] = box[i][j];
                }
            }
        }
        return res;
    }
}