问题描述
https://leetcode.com/problems/spiral-matrix/
解决思路
设置四个角，然后不断地以螺旋状将数组加入res结果集。最后需要单独处理一下最后剩下一行或者一列的情况。
代码
class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
LinkedList<Integer> res = new LinkedList<>();
if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return res;
int left = 0, right = matrix[0].length - 1;
int top = 0, bottom = matrix.length - 1;
while(left < right && top < bottom){
//in every row, from left to right
for(int i = left; i < right; i++) res.add(matrix[top][i]);
//in every coloum, from top to bottom
for(int i = top; i < bottom; i++) res.add(matrix[i][right]);
//in every row, from right to left
for(int i = right; i > left; i--) res.add(matrix[bottom][i]);
//every coloum, from bottom to top
for(int i = bottom; i > top; i--) res.add(matrix[i][left]);
left++;
right--;
top++;
bottom--;
}
//only left one coloum
if(left == right){
for(int i = top; i <= bottom; i++) res.add(matrix[i][left]);
}else if(top == bottom){//only left one row
for(int i = left; i <= right; i++) res.add(matrix[top][i]);
}
return res;
}
}
复杂度分析
- 时间复杂度：O(n)
- 空间复杂度：O(n)
