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].
打印螺旋矩阵
逐个环的打印, 对于m *n的矩阵,环的个数是 (min(n,m)+1) / 2。对于每个环顺时针打印四条边。
注意的是：最后一个环可能只包含一行或者一列数据
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class
Solution {public: vector<int> spiralOrder(vector<vector<int> > &matrix) { int
m = matrix.size(), n; if(m != 0)n = matrix[0].size(); int
cycle = m > n ? (n+1)/2 : (m+1)/2;//环的数目 vector<int>res; int
a = n, b = m;//a,b分别为当前环的宽度、高度 for(int
i = 0; i < cycle; i++, a -= 2, b -= 2) { //每个环的左上角起点是matrix[i][i],下面顺时针依次打印环的四条边 for(int
column = i; column < i+a; column++) res.push_back(matrix[i][column]); for(int
row = i+1; row < i+b; row++) res.push_back(matrix[row][i+a-1]); if(a == 1 || b == 1)break; //最后一个环只有一行或者一列 for(int
column = i+a-2; column >= i; column--) res.push_back(matrix[i+b-1][column]); for(int
row = i+b-2; row > i; row--) res.push_back(matrix[row][i]); } return
res; }}; |
Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
For example,
Given n = 3,
You should return the following matrix:
[ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ] ]
本质上和上一题是一样的，这里我们要用数字螺旋的去填充矩阵。同理，我们也是逐个环的填充，每个环顺时针逐条边填充 本文地址
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class
Solution {public: vector<vector<int> > generateMatrix(int
n) { vector<vector<int> > matrix(n, vector<int>(n)); int
a = n;//a为当前环的边长 int
val = 1; for(int
i = 0; i < n/2; i++, a -= 2) { //每个环的左上角起点是matrix[i][i],下面顺时针依次填充环的四条边 for(int
column = i; column < i+a; column++) matrix[i][column] = val++; for(int
row = i+1; row < i+a; row++) matrix[row][i+a-1] = val++; for(int
column = i+a-2; column >= i; column--) matrix[i+a-1][column] = val++; for(int
row = i+a-2; row > i; row--) matrix[row][i] = val++; } if(n % 2)matrix[n/2][n/2] = val;//n是奇数时，最后一个环只有一个数字 return
matrix; }}; |
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LeetCode:Spiral Matrix I II,布布扣,bubuko.com
原文地址：http://www.cnblogs.com/TenosDoIt/p/3774747.html
