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#yyds干货盘点# leetcode算法题：螺旋矩阵
source link: https://blog.51cto.com/u_13321676/5470405
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#yyds干货盘点# leetcode算法题：螺旋矩阵
原创给你一个 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。
输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[1,2,3,6,9,8,7,4,5]
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]
代码实现：
class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> order = new ArrayList<Integer>();
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return order;
}
int rows = matrix.length, columns = matrix[0].length;
int left = 0, right = columns - 1, top = 0, bottom = rows - 1;
while (left <= right && top <= bottom) {
for (int column = left; column <= right; column++) {
order.add(matrix[top][column]);
}
for (int row = top + 1; row <= bottom; row++) {
order.add(matrix[row][right]);
}
if (left < right && top < bottom) {
for (int column = right - 1; column > left; column--) {
order.add(matrix[bottom][column]);
}
for (int row = bottom; row > top; row--) {
order.add(matrix[row][left]);
}
}
left++;
right--;
top++;
bottom--;
}
return order;
}
}
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> order = new ArrayList<Integer>();
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return order;
}
int rows = matrix.length, columns = matrix[0].length;
int left = 0, right = columns - 1, top = 0, bottom = rows - 1;
while (left <= right && top <= bottom) {
for (int column = left; column <= right; column++) {
order.add(matrix[top][column]);
}
for (int row = top + 1; row <= bottom; row++) {
order.add(matrix[row][right]);
}
if (left < right && top < bottom) {
for (int column = right - 1; column > left; column--) {
order.add(matrix[bottom][column]);
}
for (int row = bottom; row > top; row--) {
order.add(matrix[row][left]);
}
}
left++;
right--;
top++;
bottom--;
}
return order;
}
}
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