Spiral Matrix I(II)

最新推荐文章于 2026-01-05 22:36:07 发布

原创最新推荐文章于 2026-01-05 22:36:07 发布 · 897 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #Spiral Matrix #算法

LeetCode 专栏收录该内容

112 篇文章

订阅专栏

本文详细解析了螺旋序矩阵和SpiralMatrixII算法，包括核心思路、实现步骤以及代码实现。重点突出算法的核心逻辑和边界情况处理。

题目一：螺旋序矩阵

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

注意一些边界情况。

maxlevel = min(row,col)/2+min(row,col)%2;

这个也要注意。

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> result;
        int row = matrix.size();
        if(row<=0) 
            return result;
        int col = matrix[0].size();
        if(col==1){
            for(int i=0;i<row;i++)
                result.push_back(matrix[i][0]);
            return result;            
        }
        int level = 0;
        int maxlevel = min(row,col)/2+min(row,col)%2;
        while(level<maxlevel)
        {
            for(int j=level;j<=col-1-level;j++){
                result.push_back(matrix[level][j]);
            }
            for(int i=level+1;i<row-1-level;i++){
                result.push_back(matrix[i][col-1-level]);
            }           
            if(row-1-level==level)
                break;
            for(int j=col-1-level;j>level;j--){
                result.push_back(matrix[row-1-level][j]);
            }
            for(int i=row-1-level;i>level;i--){
                result.push_back(matrix[i][level]);
            }  
            level++;
        }   
        return result;        
    }
};

还可以用递归的方法。

题目二：Spiral Matrix II

class Solution {
public:
    vector<vector<int> > generateMatrix(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        vector<vector<int>> result(n,vector<int>(n,0));  
        if(n<=0)
            return result;  
        int cur = 1;
        int level = 0;  
        int maxlevel = n/2+n%2;  
        while(level<maxlevel)  
        {  
            for(int j=level;j<=n-1-level;j++){  
                result[level][j] = cur++;  
            }  
            for(int i=level+1;i<n-1-level;i++){  
                result[i][n-1-level] = cur++;  
            }             
            if(n-1-level==level)  
                break;  
            for(int j=n-1-level;j>level;j--){  
                result[n-1-level][j] = cur++;  
            }  
            for(int i=n-1-level;i>level;i--){  
                result[i][level] = cur++;  
            }    
            level++;  
        }     
        return result;          
    }  
};