【LeetCode】Spiral Matrix

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Spiral Matrix
Total Accepted: 9568 Total Submissions: 47010 My Submissions
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].
【解题思路】
本题代码借鉴了别人的思路。
1、设定边界值，startx，endx，starty，endy。
2、左到右，上到下，右到左，下到上，直到while条件结束。

Java AC 340ms

public class Solution {    public ArrayList<Integer> spiralOrder(int[][] matrix) {ArrayList<Integer> allList = new ArrayList<Integer>();if (matrix == null || matrix.length == 0) {return allList;}allList = printfMatrix(matrix);return allList;}public ArrayList<Integer> printfMatrix(int[][] matrix) {ArrayList<Integer> allList = new ArrayList<Integer>();int m = matrix.length;int n = matrix[0].length;int startx = 0;int starty = 0;int endx = m - 1;int endy = n - 1;while (startx <= endx && starty <= endy) {for (int i = starty; i <= endy; i++) {allList.add(matrix[startx][i]);}for (int i = startx+1; i <= endx; i++) {allList.add(matrix[i][endy]);}if (startx != endx) {for (int i = endy-1; i >= starty; i--) {allList.add(matrix[endx][i]);}}if (starty != endy) {for (int i = endx-1; i > startx; i--) {allList.add(matrix[i][starty]);}}startx++;endx--;starty++;endy--;}return allList;}}

Python AC 76ms

class Solution:    # @param matrix, a list of lists of integers    # @return a list of integers    def spiralOrder(self, matrix):        list = []        if matrix is None or len(matrix) == 0:            return list        m = len(matrix)        n = len(matrix[0])        startx = 0        starty = 0        endx = m - 1        endy = n - 1        while startx <= endx and starty <= endy:            for i in range(starty, endy + 1):                list.append(matrix[startx][i])            for i in range(startx + 1, endx + 1):                list.append(matrix[i][endy])            if startx != endx:                for i in range(endy - 1, starty - 1, -1):                    list.append(matrix[endx][i])            if starty != endy:                for i in range(endx - 1, startx, -1):                    list.append(matrix[i][starty])            startx += 1            endx -= 1            starty += 1            endy -= 1        return list

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