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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?
1,O(mn),即用的一个相同大小的矩阵来记录那个位置有0.
2,O(m + n)，就是加一行一列来标记哪行哪列有0。
3,常数空间，即考虑不使用额外的空间,可以把第一行与第一列作为标记行与标记列，但是得先确定第一行与第一列本身要不要设为0。
class Solution { public: void setZeroes(vector<vector<int>>& matrix) { if (matrix.size() < 1) return; int row = matrix.size(), col = matrix[0].size(); bool r0 = false, c0 = false; for (int i = 0; i < row; ++i) { if (matrix[i][0] == 0) { c0 = true; break; } } for (int j = 0; j < col; ++j) { if (matrix[0][j] == 0) { r0 = true; break; } } for (int i = 1; i < row; ++i) { for (int j = 1; j < col; ++j) { matrix[i][0] = (matrix[i][j] == 0) ? 0 : matrix[i][0]; matrix[0][j] = (matrix[i][j] == 0) ? 0 : matrix[0][j]; } } for (int i = 1; i < row; ++i) { for (int j = 1; j < col; ++j) { matrix[i][j] = (matrix[i][0] == 0) ? 0 : matrix[i][j]; matrix[i][j] = (matrix[0][j] == 0) ? 0 : matrix[i][j]; } } for (int i = 0; i < row && c0; ++i) matrix[i][0] = 0; for (int j = 0; j < col && r0; ++j) matrix[0][j] = 0; } };
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原文地址：http://www.cnblogs.com/qiaozhoulin/p/4581259.html
