《Cracking the Coding Interview》——第1章：数组和字符串——题目7

2014-03-18 01:55

题目：给定一个MxN矩阵，如果某个元素为0，则将对应的整行和整列置为0。

解法：单独挑出一行和一列作为标记数组。因为某元素为0就全部置为0，所以不论A[i][j]为0中的j是几，第i行总会被置为0的。再用O(1)的额外空间去标记单独挑出的那一行一列是否包含0即可。要注意最后清零的顺序和范围不要错了。

代码：

 // 1.7 Write an algorithm such that if an element in an MxN matrx is 0, its antire row and column are set to 0.
 #include <cstdio>
 #include <vector>
 using namespace std;

 class Solution {
 public:
     void setMatrixZero(vector<vector<int> > &board) {
         int m, n;

         m = (int)board.size();
         if (m == ) {
             return;
         }
         n = (int)board[].size();
         if (n == ) {
             return;
         }

         int i, j;
         bool row0_has_zero = false;
         bool col0_has_zero = false;
         for (i = ; i < m; ++i) {
             if (board[i][] == ) {
                 col0_has_zero = true;
                 break;
             }
         }
         for (j = ; j < n; ++j) {
             if (board[][j] == ) {
                 row0_has_zero = true;
             }
         }
         for (i = ; i < m; ++i) {
             for (j = ; j < n; ++j) {
                 if (board[i][j] == ) {
                     board[i][] = ;
                     board[][j] = ;
                 }
             }
         }
         for (i = ; i < m; ++i) {
             if (board[i][] == ) {
                 for (j = ; j < n; ++j) {
                     board[i][j] = ;
                 }
             }
         }
         for (j = ; j < n; ++j) {
             if (board[][j] == ) {
                 for (i = ; i < m; ++i) {
                     board[i][j] = ;
                 }
             }
         }
         if (row0_has_zero) {
             for (j = ; j < n; ++j) {
                 board[][j] = ;
             }
         }
         if (col0_has_zero) {
             for (i = ; i < m; ++i) {
                 board[i][] = ;
             }
         }
     }
 };

 int main()
 {
     vector<vector<int> > board;
     int i, j;
     int m, n;
     Solution sol;

     while (scanf("%d%d", &m, &n) ==  && (m || n)) {
         board.resize(m);
         for (i = ; i < m; ++i) {
             board[i].resize(n);
         }
         for (i = ; i < m; ++i) {
             for (j = ; j < n; ++j) {
                 scanf("%d", &board[i][j]);
             }
         }
         sol.setMatrixZero(board);
         for (i = ; i < m; ++i) {
             for (j = ; j < n; ++j) {
                 if (j > ) {
                     printf(" %d", board[i][j]);
                 } else {
                     printf("%d", board[i][j]);
                 }
             }
             printf("\n");
         }
         printf("\n");
     }

     return ;
 }