剑指 Offer 29. 顺时针打印矩阵

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剑指 Offer 29. 顺时针打印矩阵

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]

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

解题思路

分四条路:

  1. 从左到右,left -> right,top > bottom,上边界下移
  2. 从上到下,top -> buttom,left > right,右边界左移
  3. 从右到左,right -> left,top > bottom,下边界上移
  4. 从下到上,buttom -> top,left > right,左边界右移

其中一条走完就可以返回结果

代码

class Solution {
    public int[] spiralOrder(int[][] matrix) {
        if (matrix == null || matrix.length == 0) return new int[0];
        int left = 0, right = matrix[0].length - 1, top = 0, bottom = matrix.length - 1, x = 0;
        int[] result = new int[(right + 1) * (bottom + 1)];
        while (true) {
            for (int i = left; i <= right; i++) result[x++] = matrix[top][i];
            if (++top > bottom) break;
            for (int i = top; i <= bottom; i++) result[x++] = matrix[i][right];
            if (left > --right) break;
            for (int i = right; i >= left; i--) result[x++] = matrix[bottom][i];
            if (top > --bottom) break;
            for (int i = bottom; i >= top; i--) result[x++] = matrix[i][left];
            if (++left > right) break;
        }
        return result;
    }
}