顺时针打印矩阵
1. 题目形容
输出一个矩阵，依照从内向里以顺时针的程序顺次打印出每一个数字
2. 示例
[[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12],
[13, 14, 15, 16]]
则顺次打印出数字1,2,3,4,8,12,16,15,14,13,9,5,6,7,11,10.
3. 解题思路
顺时针走，即向右->向下->向左->向上，一共要走（长*宽）步。
遇到边界就改变方向，当向上碰到顶的时候，四个边界都放大。思路简略，一个循环即可！
4. Java实现
import java.util.ArrayList;public class Solution { public ArrayList<Integer> printMatrix(int [][] matrix) { ArrayList<Integer> res = new ArrayList(); int rows = matrix.length; // 行的大小 int cols = matrix[0].length; // 列的大小 if (rows == 0 || cols == 0){ return res; } // 从左到右开始遍历，right最左边【列】，从上到下，bottom最底部【行】 int left = 0, right = cols-1, top = 0, bottom = rows -1; while(left <= right && top <= bottom){ // 从左到右开始遍历 for(int i = left; i <= right; i++){ res.add(matrix[top][i]); } // 从上到下开始遍历 for (int i = top+1; i <= bottom; i++){ res.add(matrix[i][right]); } // 从下往上开始遍历 if (top != bottom){ for (int i = right-1; i>= left; i--){ res.add(matrix[bottom][i]); } } //从 右往左开始遍历 if (left != right){ for (int i = bottom-1; i > top; i--){ res.add(matrix[i][left]); } } left++; right--; top++; bottom--; } return res; } }5. Python实现
# -*- coding:utf-8 -*-class Solution: # matrix类型为二维列表，须要返回列表 def printMatrix(self, matrix): if matrix == None: return rows = len(matrix) columns = len(matrix[0]) start = 0 while rows > start * 2 and columns > start * 2: self.PrintMatrixInCircle(matrix, columns, rows, start) start += 1 print('') def PrintMatrixInCircle(self, matrix, columns, rows, start): endX = columns - 1 - start endY = rows - 1 - start # 从左到右打印一行 for i in range(start, endX+1): number = matrix[start][i] print(number, ' ', end='') # 从上到下打印一行 if start < endY: for i in range(start+1, endY+1): number = matrix[i][endX] print(number, ' ', end='') # 从右到左打印一行 if start < endX and start < endY: for i in range(endX-1, start-1, -1): number = matrix[endY][i] print(number, ' ', end='') # 从下到上打印一行 if start < endX and start < endY-1: for i in range(endY-1, start, -1): number = matrix[i][start] print(number, ' ', end='')如果您感觉本文有用，请点个“在看”