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?
解题思路:
这题挺容易出问题,要考虑到如果遇到matrix[i][j]就清零第i行,第j列所有元素,会导致只要有一个零元则整个矩阵被清零。如下做法中,将零元信息记录在了第一行,第一列。但是这里有一个特殊的matrix[0][0],如果它是零,要考虑到它到底是记录的行有0,还是列有0,还是行列同时有0。
class Solution { public: void setZeroes(vector<vector<int> > &matrix) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. if(matrix.size() == 0) return; int type = 0; for(int i = 0;i < matrix.size();i++) { for(int j = 0;j < matrix[0].size();j++) { if(matrix[i][j] == 0) { matrix[i][0] = 0; matrix[0][j] = 0; if(i == 0) { if(type == 0) type = 1; else if(type == 2) type = 3; else continue; } if(j == 0) { if(type == 0) type = 2; else if(type == 1) type = 3; else continue; } } } } for(int i = 1;i < matrix.size();i++) { if(matrix[i][0] == 0) for(int j = 0;j < matrix[0].size();j++) matrix[i][j] = 0; } for(int j = 1;j < matrix[0].size();j++) { if(matrix[0][j] == 0) for(int i = 0;i < matrix.size();i++) matrix[i][j] = 0; } if(type == 1 || type == 3) for(int j = 0;j < matrix[0].size();j++) matrix[0][j] = 0; if(type == 2 || type == 3) for(int i = 0;i < matrix.size();i++) matrix[i][0] = 0; } };