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]
思路
限制边界,同时注意边界的限制条件
解答
class Solution {
public int[] spiralOrder(int[][] matrix) {
if (matrix.length == 0) return new int[0];
int left = 0, right = matrix[0].length - 1, top = 0, bottom = matrix.length - 1;
int 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;
}
}