[LeetCode]Spiral Matrix_matrix dr.liang 8.2-优快云博客

本文链接：https://blog.youkuaiyun.com/CiaoLiang/article/details/49982789

本文介绍了一种算法，用于按螺旋顺序遍历一个给定的 m x n 矩阵的所有元素，并提供了一个 C++ 实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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］手动模拟循环输出过程。

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> ret;
        if(matrix.size()==0)
            return ret;
        int colmin = 0;
        int colmax = matrix.size()-1;
        int rowmin = 0;
        int rowmax = matrix[0].size()-1;
        int i=0;
        int j=0;
        int flag = 0; //走路顺序
        while(rowmin<=rowmax&&colmin<=colmax){
            if(flag==0){
                for(j=rowmin; j<=rowmax; ++j)
                    ret.push_back(matrix[i][j]);
                j = rowmax; //漏掉这个会出错，因为j会走到rowmax+1
                ++colmin;
                flag = 1;
                continue;
            }
            if(flag == 1){
                for(i=colmin; i<=colmax; ++i)
                    ret.push_back(matrix[i][j]);
                i = colmax;
                --rowmax;
                flag = 2;
                continue;
            }
            if(flag == 2){
                for(j=rowmax; j>=rowmin;--j)
                    ret.push_back(matrix[i][j]);
                j = rowmin;
                --colmax;
                flag = 3;
                continue;
            }
            if(flag == 3){
                for(i=colmax ; i>=colmin; --i)
                    ret.push_back(matrix[i][j]);
                i = colmin;
                ++rowmin;
                flag = 0;
                continue;
            }
        }
        return ret;
    }
};