题目描述
输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字，例如，如果输入如下矩阵： 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 则依次打印出数字1,2,3,4,8,12,16,15,14,13,9,5,6,7,11,10.
python实现：
# -*- coding:utf-8 -*-
class Solution:
# matrix类型为二维列表，需要返回列表
def printMatrix(self, matrix):
# write code here
m = len(matrix)
n = len(matrix[0]) if m else 0
result = []
if m==0 or n==0:
return result
result += matrix[0]
i, j = 0, n-1#第一行最后一个元素
stepRow, stepCol = n, m#横着走和竖着走的步数
direction = 1#1(下),2(左),3(上),4(右)
while stepRow and stepCol:
if direction==1:
stepCol -= 1
for _ in range(stepCol):
i += 1
result.append(matrix[i][j])
direction = 2
elif direction==2:
stepRow -= 1
for _ in range(stepRow):
j -= 1
result.append(matrix[i][j])
direction = 3
elif direction==3:
stepCol -= 1
for _ in range(stepCol):
i -= 1
result.append(matrix[i][j])
direction = 4
elif direction==4:
stepRow -= 1
for _ in range(stepRow):
j += 1
result.append(matrix[i][j])
direction = 1
else:
pass
return result
c++实现：
class Solution {
public:
vector<int> printMatrix(vector<vector<int>> matrix) {
int m = matrix.size();
vector<int> result;
if(m==0)
return result;
int n = matrix[0].size();
if(n==0)
return result;
for(auto item: matrix[0]){
result.push_back(item);
}
int i=0, j=n-1;
int stepRow=n, stepCol=m;
int direction = 1;
while(stepRow && stepCol){
if(direction==1){
stepCol--;
for(int k=0;k<stepCol;k++){
result.push_back(matrix[++i][j]);
}
direction = 2;
}else if(direction==2){
stepRow--;
for(int k=0;k<stepRow;k++){
result.push_back(matrix[i][--j]);
}
direction = 3;
}else if(direction==3){
stepCol--;
for(int k=0;k<stepCol;k++){
result.push_back(matrix[--i][j]);
}
direction = 4;
}else if(direction==4){
stepRow--;
for(int k=0;k<stepRow;k++){
result.push_back(matrix[i][++j]);
}
direction = 1;
}//end if
}//end while
return result;
}//end printMatrix
};
转载请注明原文地址: https://ju.6miu.com/read-1200109.html
