Spiral Matrix
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
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;}
};
Spiral Matrix II
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
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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