Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example,Given the following matrix:
[[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]]
You should return [1,2,3,6,9,8,7,4,5].

Spiral Matrix I思路

1. 规律是先向右，再向下，再向左，再向上搜索，搜索的时候从matrix中取出对应位置的数放到result中，每次搜索完都需要更新rowStart || rowEnd || colStart || colEnd
2. 终止条件是：rowStart > rowEnd || colStart > colEnd

Example

rowStart->1  2  3  4
          5  6  7  8
rowEnd->  9 10 11 12
          |       |  
       colStart  colEnd
打印顺序：1234, 8 12, 11 10 9, 5, 6 7

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> result = new ArrayList<Integer>();
        
        if (matrix == null || matrix.length == 0) {
            return result;
        }
        
        int rowStart = 0;
        int rowEnd = matrix.length - 1;
        int colStart = 0;
        int colEnd = matrix[0].length - 1;
        
        //终止条件是：rowStart > rowEnd || colStart > colEnd 
        // 规律是先向右，再向下，再向左，再向上搜索，每次搜索完都需要更新rowStart || rowEnd || col Start|| colEnd
        
        while (true) {
            //->right
            for (int i = colStart; i <= colEnd; i++) {
                result.add(matrix[rowStart][i]);
            }
            rowStart++;
            if (rowStart > rowEnd) break;
            
            //->down
            for (int i = rowStart; i <= rowEnd; i++) {
                result.add(matrix[i][colEnd]);
            }
            colEnd--;
            if (colStart > colEnd) break;
            
            //->left
            for (int i = colEnd; i >= colStart; i--) {
                result.add(matrix[rowEnd][i]);
            }
            rowEnd--;
            if (rowStart > rowEnd) break;
            
            //->up
            for (int i = rowEnd; i >= rowStart; i--) {
                result.add(matrix[i][colStart]);
            }
            colStart++;
            if (colStart > colEnd) break;
        }
        return result;
    }
}

Spiral Matrix ii

Given an integer n, generate a square matrix filled with elements from 1 to n^2 in spiral order.

Example
Given n = 3,

You should return the following matrix:

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

思路

1. 与Spiral Matrix I正好相反，需要根据矩阵大小创建这个数组。同样可以用1中的规律来创建这个数组
2. 先创建一个int[n][n]大小的Matrix，因为matrix填充的数是1 - n^2，所以根据1中的规律，先向右，再向下，再向左，再向上搜索，搜索的时候将matrix中对应位置填充为cur（cur从1开始，每次填充一个数cur ++），每次搜索完都需要更新rowStart || rowEnd || colStart || colEnd
3. 终止条件是：rowStart > rowEnd || colStart > colEnd

class Solution {
    public int[][] generateMatrix(int n) {
        int[][] result = new int[n][n];
        
        if (n <= 0) {
            return result;
        }
        
        int rowStart = 0;
        int rowEnd = n - 1;
        int colStart = 0;
        int colEnd = n - 1;
        int cur = 1;
        
        while (true) {
            for (int i = colStart; i <= colEnd; i++) {
                result[rowStart][i] = cur;
                cur++;
            }
            rowStart++;
            if (rowStart > rowEnd) break;
            
            for (int i = rowStart; i <= rowEnd; i++) {
                result[i][colEnd] = cur;
                cur++;
            }
            colEnd--;
            if (colStart > colEnd) break;
            
            for (int i = colEnd; i >= colStart; i--) {
                result[rowEnd][i] = cur;
                cur++;
            }
            rowEnd--;
            
            if (rowStart > rowEnd) break;
            
            for (int i = rowEnd; i >= rowStart; i--) {
                result[i][colStart] = cur;
                cur++;
            }
            colStart++;
            if (colStart > colEnd) break;
        }
        return result;
    }
}