LeetCode

题目：

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?

思路：

1) O(mn)解法，拷贝一份matrix，然后对照这个matrix来设置原matrix的各个元素

2) O(m+n)解法，两个一维Boolean数组Rows和Columns，大小分别为m和n，如果元素（i，j）为0，则设置Rows[i]=true, Columns[j] = true。之后根据Rows和Columns中的值来对每行和每列进行设置为0.

3) 在方法二的基础上，借用matrix的第一行和第一列来表示相应的列和行是否为0，但这个时候需要对第一行和第一列进行特殊化处理，我们就用两个Boolean变量来应对。

package array;

public class SetMatrixZeroes {

    public void setZeroes(int[][] matrix) {
        boolean firstRow = false;
        boolean firstColumn = false;
        int m = matrix.length;
        int n = matrix[0].length;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == 0) {
                    if (i == 0) firstRow = true;
                    if (j == 0) firstColumn = true;
                    matrix[i][0] = matrix[0][j] = 0;
                }
            }
        }
        
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                if (matrix[i][0] == 0 || matrix[0][j] == 0)
                    matrix[i][j] = 0;
            }
        }
        
        if (firstRow) {
            for (int i = 0; i < n; ++i)
                matrix[0][i] = 0;
        }
        
        if (firstColumn) {
            for (int i = 0; i < m; ++i)
                matrix[i][0] = 0;
        }
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int[][] matrix = { { 1, 0, 1 }, { 1, 1, 1 }, { 0, 1, 1 } };
        SetMatrixZeroes s = new SetMatrixZeroes();
        s.setZeroes(matrix);
    }

}