Title Description

To give you one m That's ok n Columns of the matrix matrix , Please follow Clockwise spiral sequence , Returns all elements in the matrix .
Example 1 ：

 Input ：matrix = [[1,2,3],[4,5,6],[7,8,9]]
 Output ：[1,2,3,6,9,8,7,4,5]

Example 2 ：

 Input ：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
 Output ：[1,2,3,4,8,12,11,10,9,5,6,7]

Ideas

Separate use up、down、left、right To represent the boundary of the matrix

 The order of traversal 
1. Left to right    matrix[up][left]....matrix[up][right]
2. Up and down    matrix[up+1][right]....matrix[down][right] 			// Be careful up+1 For the indentation of the boundary 
3. Right to left    matrix[down][right-1]....matrix[down][left] 		// Be careful right-1 For the indentation of the boundary 
4. Down to up    matrix[down+1][left]....matrix[up][left] 			//down-1 ditto 

The main difficulty of this problem is how to determine the four boundaries , When initializing ,up Namely 0,down yes x_len-1,left yes 0,right yes y_len-1. Enter... In the above traversal order while loop , After each traversal, you need to Boundary indent and Boundary judgment , if left>right perhaps up>down when , Indicates the end of the traversal , direct break end . The core of the following code is that after each operation ( Left to right 、 Up and down 、 Right to left and bottom to top ) after , Will indent and judge the boundary .

vector<int> spiralOrder(vector<vector<int>>& matrix) {
    
	vector<int> ans;
	int x_len=matrix.size();// Row number 
	int y_len=matrix[0].size();// Number of columns 
	// Mark the boundaries of all four points 
	int left=0,right=y_len-1,up=0,down=x_len-1;
	while(true){
    
	
		// From left to right   assignment 
		for(int i=left;i<=right;i++)
			ans.push_back(matrix[up][i]);
		if(++up > down)// Indent the boundary  up=up+1 operation   Pay attention to ++up  stay up==down  It was all about up=up+1 Replacement of 
			break;
			
		// From top to bottom   At this time up The value of is compared to the previous for Medium up We've added 1
		for(int i=up;i<=down;i++)
			ans.push_back(matrix[i][right]);
		if(--right < left)// Indent the boundary  right=right-1 operation 
			break;
			
		// From right to left   Be careful right Change in value of 
		for(int i=right;i>=left;i--)
			ans.push_back(matrix[down][i]);
		if(--down < up)// Indent the boundary  down=down-1 operation 
			break;
			
		// From bottom to top   Be careful down Change in value of 
		for(int i=down;i>=up;i--)
			ans.push_back(matrix[i][left]);
		if(++left > right)// Indent the boundary  left=left+1 operation 
			break;
	}
	return ans;
}

Reference resources
Comment area Anatole Big man's answer