in this problem, both monish001 and hawstein use O(n^2) solution which first tranverse the char** a to find the a[i][j]==0 and mark that row/col to be: TRUE.
Then the send loop make the TURE row/col to be all 0.
//https://github.com/monish001/CPP-Programs/blob/master/CTCI/1.7Zeros.cpp
//Write an algorithm such that if an element in an MxN matrix is 0, its entire row and column is set to 0.
#include
#include
using namespace std;
void swap(char &a, char&b){
std::swap(a,b);
// cout< r, c;
for(int i=0; i
本文介绍了一种处理矩阵中零元素的算法,当发现矩阵中的某个元素为零时,将其所在行与列的所有元素置零。文章提供了一个O(n^2)的解决方案,并附带了详细的C++实现代码。
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