L54. Spiral Matrix

bfs

Problem:

螺旋打印

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

Example 1:

Input:
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]

Example 2:

Input:
[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

Solution：

• 旋转打印

public List<Integer> spiralOrder(int[][] matrix) {
    List<Integer> res = new LinkedList<>();
    if(matrix == null || matrix.length == 0) return res;
    int n = matrix.length, m = matrix[0].length;
    int up = 0, down = n - 1;
    int left = 0, right = m - 1;
    while(res.size() <  n * m){
        for(int j = left; j <= right && res.size() < n*m; j++){
            res.add(matrix[up][j]);
        }
        for(int i = up + 1; i <= down-1 && res.size() < n*m; i++){
            res.add(matrix[i][right]);
        }
        for(int j = right; j >= left && res.size() < n*m; j--){
            res.add(matrix[down][j]);
        }
        for(int i = down - 1; i >= up+1 && res.size() < n*m; i--){
            res.add(matrix[i][left]);
        }
        left++; right--; up++; down--;
    }
    return res;
}