剑指 Offer 29. 顺时针打印矩阵【力扣】

题意理解

给定一个矩阵，按照棒棒糖顺时针旋转，得到一个数组，打印该数组

问题分析

设置一个访问过数组，一个大循环内套4个小循环，分别是上，右，下，左。大循环条件是长度等于矩阵元素数量。

其他

力扣

这个就是模拟，搞清楚就可以了。

链接

class Solution {
public:
    vector spiralOrder(vector>& matrix) {
        vector res;
        int m = matrix.size();
        if (m == 0) return res;
        int n = matrix[0].size();
        int max_len = m * n;
        vector> visited(m, vector(n));
        int len = 0;
        int i = 0, j = 0;
        //cout << "start" << endl;
        while(len < max_len) {
            //cout << "loop " << endl;
            for (j = j; j < n && visited[i][j] == 0; j++) {
                //cout << "eee" << endl;
                //cout << "上 i j val" << 't' << i << 't' << j << 't' << matrix[i][j] << endl;
                res.push_back(matrix[i][j]);
                visited[i][j] = 1;
                len++;
            }
            i++; j--;
            for (i = i; i < m && visited[i][j] == 0; i++) {
                //cout << "右 i j val" << 't' << i << 't' << j << 't' << matrix[i][j] << endl;                
                res.push_back(matrix[i][j]);
                visited[i][j] = 1;
                len++;
            }
            j--; i--;
            for(j = j ; j >= 0 && visited[i][j] == 0; j--) {
                //cout << "下 i j val" << 't' << i << 't' << j << 't' << matrix[i][j] << endl;                
                res.push_back(matrix[i][j]);
                visited[i][j] = 1;
                len++;
            }
            i--; j++;
            //cout << "左 " << i << 't' << j << endl;
            for(i = i; i >= 0 && visited[i][j] == 0; i--) {
                //cout << "左 i j val" << 't' << i << 't' << j << 't' << matrix[i][j] << endl;
                res.push_back(matrix[i][j]);
                visited[i][j] = 1;
                len++;
            }
            i++;j++;
        }
        return res;
    }
};