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package com.yangchd.leetcode.medium;

import java.util.ArrayList;

import java.util.List;

/**

* @author yangchd 2018/10/11.

*

* 54. Spiral Matrix

* 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]

*

*

*/

public class SpiralMatrix {

class Solution {

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

List<Integer> list = new ArrayList<Integer>();

addInteger(matrix, 0, list);

return list;

}

private void addInteger(int[][] matrix, int time, List<Integer> list) {

int high = matrix.length - 2 * time;

if (high <= 0) {

return;

}

int width = matrix[0].length - 2 * time;

if (width <= 0) {

return;

}

if (high == 1) {

for (int line = 0; line < width; line++) {

list.add(matrix[time][time + line]);

}

return;

}

if (width == 1) {

for (int col = 0; col < high; col++) {

list.add(matrix[time + col][time]);

}

return;

}

// 上边

for (int top = 0; top < width - 1; top++) {

list.add(matrix[time][top + time]);

}

// 右边

for (int right = 0; right < high - 1; right++) {

list.add(matrix[right + time][width + time - 1]);

}

// 下边

for (int bottom = time + width - 1; bottom > time; bottom--) {

list.add(matrix[high + time - 1][bottom]);

}

// 左边

for (int left = high + time - 1; left > time; left--) {

list.add(matrix[left][time]);

}

addInteger(matrix, time + 1, list);

}

}

}