[LeetCode97]Set Matrix Zeroes

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本文介绍了一种高效的矩阵置零算法，当矩阵中的元素为0时，将该元素所在的整行和整列置零。提供了两种解决方案，一种使用O(m+n)额外空间，另一种采用常数空间复杂度的方法。后者巧妙地利用了矩阵本身的第一行和第一列来记录哪些行和列需要被置零。

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

click to show follow up.

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?

Analysis:

The solution using O(mn) space is easy, create another flag matrix for each position.

The solution using O(m+n), create two arrays to store specific row and column zero info.

The solution using constant space need us to multiplex use the matrix itself.

Steps:

1. Determine the first row and column is 0 or not.

2. Use matrix first row and column to store 0 information

3. Set matrix 0 according to the first row and column

4. Set first row and column to 0, according to step 1.

Java

public void setZeroes(int[][] matrix) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        boolean firstRow = false, firstColumn = false;
        for(int i=0; i<matrix.length; i++){
        	if(matrix[i][0] == 0){
        		firstColumn = true;
        		break;
        	}
        }
        
        for(int i=0; i<matrix[0].length;i++){
        	if(matrix[0][i] == 0){
        		firstRow = true;
        		break;
        	}
        }
        
        for(int i=1; i<matrix.length;i++){
        	for(int j=1; j<matrix[0].length;j++){
        		if(matrix[i][j] == 0){
        			matrix[i][0] = 0;
        			matrix[0][j] = 0;
        		}
        	}
        }
        
        for(int i=1; i<matrix.length;i++){
        	for(int j=1; j<matrix[0].length;j++){
        		if(matrix[i][0] == 0 || matrix[0][j]==0){
        			matrix[i][j] = 0;
        		}
        	}
        }
        
        if(firstColumn){
        	for(int i=0; i<matrix.length;i++)
        		matrix[i][0] =0;
        }
        
        if(firstRow){
        	for(int i=0; i<matrix[0].length;i++)
        		matrix[0][i] =0;
        }
    }

c++

void setZeroes(vector<vector<int> > &matrix) {
        assert(matrix.size() >0);
    int row = matrix.size();
    int column = matrix[0].size();
    bool zeroRow = false, zeroCol = false;
    for(int i=0; i<column; i++){
        if(matrix[0][i] == 0)
            zeroRow = true;
    }
    for(int i=0; i<row; i++){
        if(matrix[i][0] == 0)
            zeroCol = true;
    }
    for(int i=1; i<row; i++){
        for(int j=1; j<column; j++){
            if(matrix[i][j]==0){
                matrix[i][0] = 0;
                matrix[0][j] = 0;
            }
        }
    }
    for(int i=1; i<row; i++){
        for(int j=1; j<column; j++){
            if(matrix[i][0]==0 || matrix[0][j]==0)
                matrix[i][j] = 0;
        }
    }
    if(zeroRow == true){
        for(int i=0; i<column;i++){
            matrix[0][i] = 0;
        }
    }
    if(zeroCol == true){
        for(int i=0; i<row;i++){
            matrix[i][0] = 0;
        }
    }
    }