LeetCode | 54. 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].

思路：蛇形矩阵。设立四个方向的边界，然后模拟一遍即可。

// 3 msclass Solution {public:/*蛇形矩阵 [   [ 1, 2, 3 ],   [ 4, 5, 6 ],   [ 7, 8, 9 ] ] => [1,2,3,6,9,8,7,4,5]*/    vector<int> spiralOrder(vector<vector<int> >& matrix)    {        vector<int> res;        int x = 0, y = 0, tot = 0;  //tot用于计数，作为停止条件        int m = matrix.size();        if(m == 0)      //错误输入数据，特殊处理            return res;        int n = matrix[0].size();        int up = 0, down = m-1, left = 0, right = n-1;        if(n==0)        //错误输入数据，特殊处理            return res;        //开始遍历        while(tot < m*n)        {            while(x==up && tot < m*n)     //最上面一行            {                res.push_back(matrix[x][y]);                tot++;                if(y == right)                    x++;                else                    y++;            }            up += 1;            while(y==right  && tot < m*n)            {                res.push_back(matrix[x][y]);                tot++;                if(x == down)                    y--;                else                    x++;            }            right -= 1;            while(x==down && tot < m*n)            {                res.push_back(matrix[x][y]);                tot++;                if(y == left)                    x--;                else                    y--;            }            down -= 1;            while(y==left && tot < m*n)            {                res.push_back(matrix[x][y]);                tot++;                if(x == up)                    y++;                else                    x--;            }            left += 1;        }        //cout<<up<<down<<left<<right;        return res;    }};

在solution中看到更简洁的代码（思路是一样的），据称 0ms，交上去呵呵了，也是3 ms

class Solution {public:    vector<int> spiralOrder(vector<vector<int>>& matrix) {        if (matrix.empty()) return {};        int m = matrix.size(), n = matrix[0].size();        vector<int> spiral(m * n);        int u = 0, d = m - 1, l = 0, r = n - 1, k = 0;        while (true) {            // up            for (int col = l; col <= r; col++) spiral[k++] = matrix[u][col];            if (++u > d) break;            // right            for (int row = u; row <= d; row++) spiral[k++] = matrix[row][r];            if (--r < l) break;            // down            for (int col = r; col >= l; col--) spiral[k++] = matrix[d][col];            if (--d < u) break;            // left            for (int row = d; row >= u; row--) spiral[k++] = matrix[row][l];            if (++l > r) break;        }        return spiral;    }};