[LeetCode] 100: Set Matrix Zeroes

[Problem]

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?

[Solution]

class Solution {
public:
void setZeroes(vector<vector<int> > &matrix) {
// Note: The Solution object is instantiated only once and is reused by each test case.

// invalid
if(matrix.size() == 0 || matrix[0].size() == 0)return;

// scan the first row and the first column
bool firstRow = false, firstColumn = false;// if the firstRow/firstColumn contains 0
for(int i = 0; i < matrix.size(); ++i){
if(matrix[i][0] == 0){
firstColumn = true;
break;
}
}
for(int i = 0; i < matrix[0].size(); ++i){
if(matrix[0][i] == 0){
firstRow = true;
break;
}
}

// scan the remain
for(int i = 1; i < matrix.size(); ++i){
for(int j = 1; j < matrix[i].size(); ++j){
if(matrix[i][j] == 0){
matrix[0][j] = 0;
matrix[i][0] = 0;
}
}
}

// set zeroes
for(int i = 1; i < matrix.size(); ++i){
if(matrix[i][0] == 0){
for(int j = 1; j < matrix[i].size(); ++j){
matrix[i][j] = 0;
}
}
}
for(int i = 1; i < matrix[0].size(); ++i){
if(matrix[0][i] == 0){
for(int j = 1; j < matrix.size(); ++j){
matrix[j][i] = 0;
}
}
}

// set the first row and the first column
if(firstRow){
for(int i = 0; i < matrix[0].size(); ++i){
matrix[0][i] = 0;
}
}
if(firstColumn){
for(int i = 0; i < matrix.size(); ++i){
matrix[i][0] = 0;
}
}
}
};

说明：版权所有，转载请注明出处。Coder007的博客