54. Spiral Matrix

题目：

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].、

题意：

给定一个m行n列的数组，将该数组按螺旋状输出。

思路：

转载地址：https://leetcode.com/discuss/38974/a-concise-c-implementation-based-on-directions

When traversing the matrix in the spiral order, at any time we follow one out of the following four directions: RIGHT DOWN LEFT UP. Suppose we are working on a 5 x 3 matrix as such:

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Imagine a cursor starts off at (0, -1), i.e. the position at '0', then we can achieve the spiral order by doing the following:

1. Go right 5 times
2. Go down 2 times
3. Go left 4 times
4. Go up 1 times.
5. Go right 3 times
6. Go down 0 times -> quit

Notice that the directions we choose always follow the order 'right->down->left->up', and for horizontal movements, the number of shifts follows:{5, 4, 3}, and vertical movements follows {2, 1, 0}.

Thus, we can make use of a direction matrix that records the offset for all directions, then an array of two elements that stores the number of shifts for horizontal and vertical movements, respectively. This way, we really just need one for loop instead of four.

Another good thing about this implementation is that: If later we decided to do spiral traversal on a different direction (e.g. Counterclockwise), then we only need to change the Direction matrix; the main loop does not need to be touched.

代码：

class Solution {public:    vector<int> spiralOrder(vector<vector<int>>& matrix) {            vector<vector<int> > dirs{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};    vector<int> res;        int nr = matrix.size();         if (nr == 0)         return res;        int nc = matrix[0].size();      if (nc == 0)         return res;    vector<int> nSteps{nc, nr-1};    int iDir = 0;   // index of direction.    int ir = 0, ic = -1;    // initial position    while (nSteps[iDir%2]) {        for (int i = 0; i < nSteps[iDir%2]; ++i) {            ir += dirs[iDir][0];             ic += dirs[iDir][1];            res.push_back(matrix[ir][ic]);        }        nSteps[iDir%2]--;        iDir = (iDir + 1) % 4;    }    return res;    }};