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1. 题目信息

给定一个正整数 n，生成一个包含 1 到 n2 所有元素，且元素按顺时针顺序螺旋排列的正方形矩阵。

示例:输入: 3输出:[ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ]]

来源：力扣（LeetCode）

2. LeetCode 59 解题

类似题目：

• 创建变量top、bottom表示上下行的区间，left、right表示列的区间
  

class Solution {
   public:    vector
   
    
     > generateMatrix(int n) {
           vector
     
      
        > v(n, vector
       
        (n,0));        if(n == 0)         	return {
   };        int i, num = 0, total = n*n, top = 0, bottom = n-1, left = 0, right = n-1;        while(num < total)        {
           	for(i = left; i <= right; ++i)        		v[top][i] = ++num;        	++top;        	for(i = top; i <= bottom; ++i)        		v[i][right] = ++num;        	--right;        	for(i = right; i >= left; --i)        		v[bottom][i] = ++num;        	--bottom;        	for(i = bottom; i >= top; --i)        		v[i][left] = ++num;        	++left;        }        return v;    }};
       
      
     
    
   

3. LeetCode 54. 螺旋矩阵

给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。

class Solution {
   public:    vector
   
     spiralOrder(vector
    
     
      >& matrix) {
           if(matrix.empty())         	return {
   };        vector
      
        ans;        int m = matrix.size(), n = matrix[0].size();        int i, num = 0, total = m*n, top = 0, bottom = m-1, left = 0, right = n-1;        while(num < total)        {
           	for(i = left; i <= right; ++i)        		ans.push_back(matrix[top][i]),num++;            if(num == total)                break;        	++top;        	for(i = top; i <= bottom; ++i)        		ans.push_back(matrix[i][right]),num++;            if(num == total)                break;        	--right;        	for(i = right; i >= left; --i)        		ans.push_back(matrix[bottom][i]),num++;            if(num == total)                break;        	--bottom;        	for(i = bottom; i >= top; --i)        		ans.push_back(matrix[i][left]),num++;            if(num == total)                break;        	++left;        }        return ans;    }};
      
     
    
   

4.《剑指Offer》面试题29

class Solution {
   	//2020.2.22public:    vector
   
     spiralOrder(vector
    
     
      >& matrix) {
           if(matrix.size()==0 || matrix[0].size()==0)        	return {
   };        int i = 0, k = 0, count = matrix.size()*matrix[0].size();        int left = 0, right = matrix[0].size()-1, up = 0, bottom = matrix.size()-1;        vector
      
        ans(count);        while(count)        {
           	i = left;        	while(count && i <= right)        	{
           		ans[k++] = matrix[up][i++];        		count--;        	}        	up++,        	i = up;        	while(count && i <= bottom)        	{
           		ans[k++] = matrix[i++][right];        		count--;        	}        	right--;        	i = right;        	while(count && i >= left)        	{
           		ans[k++] = matrix[bottom][i--];        		count--;        	}        	bottom--;        	i = bottom;        	while(count && i >= up)        	{
           		ans[k++] = matrix[i--][left];        		count--;        	}            left++;        }        return ans;    }};
      
     
    
   

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