54. 螺旋矩阵
Difficulty: 中等
给你一个 m 行 n 列的矩阵 matrix ,请按照 顺时针螺旋顺序 ,返回矩阵中的所有元素。
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]
提示:
m == matrix.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100
Solution
四个指针来进行定位遍历,top,right,bottom,right
- left—right,top++
- top—bottom right++
- right—left bottom—
- bottom—top left++
特别要注意的是,在一次循环里,如果矩阵是奇数行、偶数列的时候有问题,会出错,这时要在循环体内多判断退出条件
class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
if (matrix == null) {
return null;
}
if (matrix.length == 0) {
return new ArrayList<>();
}
int top = 0;
int bottom = matrix.length - 1;
int left = 0;
int right = matrix[0].length - 1;
List<Integer> list = new ArrayList<>();
while (left <= right && top <= bottom) {
for (int i = left; i <= right; i++) {
list.add(matrix[top][i]);
}
top++;
for (int i = top; i <= bottom; i++) {
list.add(matrix[i][right]);
}
right--;
if (left > right || top > bottom) {
break;
}
for (int i = right; i >= left; i--) {
list.add(matrix[bottom][i]);
}
bottom--;
for (int i = bottom; i >= top; i--) {
list.add(matrix[i][left]);
}
left++;
}
return list;
}
}
