The sword refers to Offer 29 (array 5). Print the matrix clockwise
Problem Description:
Enter a matrix and print each number in clockwise order from the outside to the inside.
example：
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]
Problem-solving ideas:Idea link
Consider setting the four boundaries of the matrix "left, top, right, and bottom" to simulate the above matrix traversal order.
The matrix has four boundaries: top, bottom, left, and right. Every time the boundary is traversed, the direction is changed, and the boundary is tightened.
There are four traversal directions in total:
- Left -> Right, res.append(matrix[top][i]), after traversing to the boundary, top moves down;
- Up -> Down, res.append(matrix[i][right]), after traversing to the boundary, right moves to the left;
- Right -> Left, res.append(matrix[bottom][i]), after traversing to the boundary, the bottom moves up;
- Down -> Up, res.append(matrix[i][left]), after traversing to the boundary, left moves to the right;
Implemented using recursive code:
class Solution {
public int[] spiralOrder(int[][] matrix) {
//判断矩阵是否为空
if(matrix.length == 0 || matrix == null){
return new int[0];
}
//根据顺时针的运动方式建立初始指针
int n = matrix.length,m = matrix[0].length;
int left = 0,top = 0,right = m - 1,bottom = n - 1;
//建立返回的数组
int[] number = new int[m * n];
int index = 0;
while(true){
//从矩阵位置（0,0）开始从左到右
for(int i = left;i <= right;i++){
number[index++] = matrix[top][i];
}
//遍历完此行top值下移，直到大于bottom值
top++;
if(top > bottom){
break;
}
//从上到下
for(int i = top;i <= bottom;i++){
number[index++] = matrix[i][right];
}
right--;
if(right < left){
break;
}
//从右到左
for(int i = right;i >= left;i--){
number[index++] = matrix[bottom][i];
}
bottom--;
if(bottom < top){
break;
}
//从下到上
for(int i = bottom;i >= top;i--){
number[index++] = matrix[i][left];
}
left++;
if(left > right){
break;
}
}
return number;
}
}