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].
每层以(i,i)位置出发，向右到达(i,n-1-i)，向下到达(m-1-i,n-1-i)，向左到达(m-1-i,i)，再向上回到起点。
所有层次遍历完成后，即得到所求数组
需要注意的是，当只剩一行或者一列，也就是i==m-1-i或者i==n-1-i时，不要来回重复遍历。
class Solution { public: vector<int> spiralOrder(vector<vector<int> > &matrix) { vector<int> result; if(matrix.empty() || matrix[0].empty()) return result; int m = matrix.size(); int n = matrix[0].size(); int minSize = min(m,n); int level = (minSize%2==0)?(minSize/2):(minSize/2+1); for(int i = 0; i < level; i ++) {//start with matrix[i][i] //from up-left to up-right, row i for(int j = i; j < n-i; j ++) result.push_back(matrix[i][j]); //from up-right to down-right, col n-1-i for(int j = i+1; j < m-i; j ++) result.push_back(matrix[j][n-1-i]); //from down-right to down-left, row m-1-i if(m-1-i != i) {//last row for(int j = n-1-i-1; j >= i; j --) result.push_back(matrix[m-1-i][j]); } //from down-left to up-left if(i != n-1-i) {//last col for(int j = m-1-i-1; j > i; j --) result.push_back(matrix[j][i]); } } return result; } };
原文：http://www.cnblogs.com/ganganloveu/p/4157376.html