SpiralMatrix

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54.problem

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

Example 1:

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

Example 2:

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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]

key

很简单的循环输出的例子，从【0,0】的位置顺时针扫一圈，然后缩小一圈，继续扫描，不过有一个细节就是第三次第四循环前，要判断一下,防止最后一层循环只有一行

solution

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public List<Integer> spiralOrder(int[][] matrix) {

		List<Integer> ans = new ArrayList<Integer>();
		if (matrix.length == 0)
			return ans;
		int rs = 0, re = matrix.length - 1;// rowStart rowEnd
		int cs = 0, ce = matrix[0].length - 1;// columnStart columnEnd
		while (rs <= re && cs <= ce) {
			for (int i = cs; i <= ce; i++) {
				ans.add(matrix[rs][i]);
			}
			for(int j=rs+1;j<=re;j++) {
				ans.add(matrix[j][ce]);
			}
			if(rs<re&&cs<ce) {
				for(int k=ce-1;k>cs;k--) {
					ans.add(matrix[re][k]);
				}
				for(int l=re;l>rs;l--) {
					ans.add(matrix[l][cs]);
				}
			}
			
			rs++;
			re--;
			cs++;
			ce--;
		}
		return ans;
	}

perfect

1
yehh,I'm the perfect