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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.
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); } }
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原文地址：http://www.cnblogs.com/shuaiwhu/p/5093022.html
