题目要求

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].

分析

举个例子自己从头到尾把数字列出来,很容易就找到规律了:
假设一维数组的坐标为x,取值范围是xMin~xMax;二维数组的坐标为y,取值范围是yMin~yMax。(也就是数组表示为int[y][x])
1. 从左到右,y=yMin,x: xMin->xMax,yMin++
2. 从上到下,x=xMax,y: yMin->yMax,xMax--
3. 从右到左,y=yMax,x: xMax->xMin,yMax--
4. 从下到上,x=xMin,y: yMax->uMin,xMin++
结束条件,xMin==xMax或者yMin==yMax

还要要注意的地方:空数组的情况要处理。

Java代码

public static ArrayList<Integer> spiralOrder(int[][] matrix) {
ArrayList<Integer> order = new ArrayList<Integer>();
if (matrix.length == 0 || matrix[0].length == 0) return order; int xMin = 0;
int yMin = 0;
int xMax = matrix[0].length - 1;
int yMax = matrix.length - 1; order.add(matrix[0][0]); int i = 0, j = 0;
while (true) {
while (i < xMax) order.add(matrix[j][++i]);
if (++yMin > yMax) break; while (j < yMax) order.add(matrix[++j][i]);
if (xMin > --xMax) break; while (i > xMin) order.add(matrix[j][--i]);
if (yMin > --yMax) break; while (j > yMin) order.add(matrix[--j][i]);
if (++xMin > xMax) break;
}
return order;
}

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