[LeetCode] Spiral Matrix、Rotate Image、Spiral Matrix II、Set Matrix Zeroes、Search a 2D Matrix

Spiral Matrix：

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].

一圈一圈的加，注意下标就好了。

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int> > &mat) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        
		if(mat.empty())
				return vector<int>();
		int height=mat.size(),width=mat[0].size();
		int x=0;
		vector<int> seq;
		while(height>1&&width>1)
		{
			for(int i=0;i<width-1;i++)
				seq.push_back(mat[x][x+i]);
			for(int i=0;i<height-1;i++)
				seq.push_back(mat[x+i][x+width-1]);
			for(int i=0;i<width-1;i++)
				seq.push_back(mat[x+height-1][x+width-1-i]);
			for(int i=0;i<height-1;i++)
				seq.push_back(mat[x+height-1-i][x]);
			x+=1;
			width-=2;
			height-=2;
		}
		if(height==1)
			for(int i=0;i<width;i++)
				seq.push_back(mat[x][x+i]);
		else if(width==1)
			for(int i=0;i<height;i++)
				seq.push_back(mat[x+i][x]);
		return seq;
    }
};

Rotate Image:

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Follow up:
Could you do this in-place?

也是一样，一圈一圈地转。

class Solution {
public:
	void rotate(vector<vector<int> > &mat) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function

		int n=mat.size();
		int stx=0,sty=0;
		int w=n,h=n;

		while(w>=1&&h>=1)
		{
			for(int i=0;i<w-1;i++)
			{
				int tmp=mat[stx][sty+i];
				swap(mat[stx+h-1-i][sty],mat[stx][sty+i]);
				swap(mat[stx+h-1][sty+w-1-i],mat[stx+h-1-i][sty]);
				swap(mat[stx+i][sty+w-1],mat[stx+h-1][sty+w-1-i]);
				mat[stx+i][sty+w-1]=tmp;
			}
			stx+=1;
			sty+=1;
			w-=2;
			h-=2;
		}
	}
};

Spiral Matrix II：

Given an integer n, generate a square matrix filled with elements from 1 to n² in spiral order.

For example,
Given n = 3,

You should return the following matrix:

[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

跟第一个是一样的，我把一圈圈遍历时每次的范围改了下，这回要清楚一些。

class Solution {
public:
    vector<vector<int> > generateMatrix(int n) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function

	//	assert(n>=0);
		if ( n==0 )
			return vector<vector<int> >();
		vector<vector<int> > mat(n,vector<int>(n,0));
		int x=0,y=0;
		int width=n;
		int k=1;
		while(width>0)
		{
			if ( width==1 )
			{
				mat[x][y]=k++;
				break;
			}
			int tWidth=width-1;
			for(int i=0;i<tWidth;i++)
				mat[x][y+i]=k++;
			for(int i=0;i<tWidth;i++)
				mat[x+i][y+width-1]=k++;
			for(int i=0;i<tWidth;i++)
				mat[x+width-1][y+width-1-i]=k++;
			for(int i=0;i<tWidth;i++)
				mat[x+width-1-i][y]=k++;
			width-=2;
			x++,y++;
		}
		return mat;
	}
};

Set Matrix Zeroes：

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.

Follow up:

Did you use extra space?
A straight forward solution using O(mn) space is probably a bad idea.
A simple improvement uses O(m + n) space, but still not the best solution.
Could you devise a constant space solution?

class Solution {
public:
    void setZeroes(vector<vector<int> > &mat) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function

		int m=mat.size();
		int n=mat[0].size();

		bool setCol=false,setRow=false;

		if ( mat[0][0]==0 )
		{
			setCol=setRow=true;
		}
		else
		{
			for(int i=0;i<m;i++)
				if ( mat[i][0]==0 )
					setCol=true;
			for(int j=0;j<n;j++)
				if ( mat[0][j]==0 )
					setRow=true;
		}
		for(int i=1;i<m;i++)
			for(int j=1;j<n;j++)
			{
				if ( mat[i][j]==0 )
					mat[i][0]=mat[0][j]=0;
			}
		for(int i=1;i<m;i++)
			for(int j=1;j<n;j++)
				if (mat[i][0]==0||mat[0][j]==0)
					mat[i][j]=0;
		if ( setRow )
			for(int j=0;j<n;j++)
				mat[0][j]=0;
		if ( setCol )
			for(int i=0;i<m;i++)
				mat[i][0]=0;
	}
};

Search a 2D Matrix：

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

• Integers in each row are sorted from left to right.
• The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &mat, int t) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        int m=mat.size();
	    if ( m==0 )
		    return false;
	    int n= mat[0].size();
	    int x=0,y=n-1;
	    while(x<m&&y>=0)
	    {
		    if ( mat[x][y]==t)
			    return true;
		    else if ( mat[x][y]<t )
			    x++;
		    else
			    y--;
	    }
	    return false;
    }
};