Leetcode#54 Spiral Matrix

简单模拟，用t,b,l,r分别表示当前的上下左右四个边界，然后遍历即可

代码：

 1 vector<int> spiralOrder(vector<vector<int> > &matrix) {
 2   if (matrix.empty() || matrix[0].empty()) return vector<int>();
 3 
 4   vector<int> path;
 5   int m = matrix.size();
 6   int n = matrix[0].size();
 7   int t = 0;
 8   int b = m - 1;
 9   int l = 0;
10   int r = n - 1;
11   int i = 0;
12   int j = 0;
13 
14   while (t <= b && l <= r) {
15     while (t <= b && l <= r && j <= r) path.push_back(matrix[i][j++]);
16     i++;
17     j--;
18     t++;
19     while (t <= b && l <= r && i <= b) path.push_back(matrix[i++][j]);
20     i--;
21     j--;
22     r--;
23     while (t <= b && l <= r && j >= l) path.push_back(matrix[i][j--]);
24     i--;
25     j++;
26     b--;
27     while (t <= b && l <= r && i >= t) path.push_back(matrix[i--][j]);
28     i++;
29     j++;
30     l++;
31   }
32 
33   return path;
34 }