05 August 2008
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]
要求由外层向内层螺旋打印数组，只能一行一列地打印，先往右，再往下，再往左，最后往上，用四个变量来记录打印的位置，下一轮从新的打印位置开始。
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class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> result = new ArrayList<>();
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return result;
int rowBegin = 0;
int colEnd = matrix[0].length - 1;
int colBegin = 0;
int rowEnd = matrix.length - 1;
// 这里包含等于，所以下面三四两个循环要if检查一下，防止打印两遍，可以以一行或一列来理解
while (rowBegin <= rowEnd && colBegin <= colEnd) {
//从左向右
for (int i = colBegin; i <= colEnd; i++) {
result.add(matrix[rowBegin][i]);
}
rowBegin++;
//从上到下，这里rowBegin在上面已经+1了，所以角落的元素不会重复打印，以下类似
for (int i = rowBegin; i <= rowEnd; i++) {
result.add(matrix[i][colEnd]);
}
colEnd--;
//从右到左
if (rowBegin <= rowEnd) {//防止行重复打印，可能在上面“从左到右”打印过了
for (int i = colEnd; i >= colBegin; i--) {
result.add(matrix[rowEnd][i]);
}
}
rowEnd--;
//从下到上
if (colBegin <= colEnd) {//防止列重复打印，可能在上面“从上到下”打印过了
for (int i = rowEnd; i >= rowBegin; i--) {
result.add(matrix[i][colBegin]);
}
}
colBegin++;
}
return result;
}
}