《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 // 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.

  2 #include <cstdio>

  3 #include <vector>

  4 using namespace std;

  5 

  6 class Solution {

  7 public:

  8     void setMatrixZero(vector<vector<int> > &board) {

  9         int m, n;

 10         

 11         m = (int)board.size();

 12         if (m == 0) {

 13             return;

 14         }

 15         n = (int)board[0].size();

 16         if (n == 0) {

 17             return;

 18         }

 19         

 20         int i, j;

 21         bool row0_has_zero = false;

 22         bool col0_has_zero = false;

 23         for (i = 0; i < m; ++i) {

 24             if (board[i][0] == 0) {

 25                 col0_has_zero = true;

 26                 break;

 27             }

 28         }

 29         for (j = 0; j < n; ++j) {

 30             if (board[0][j] == 0) {

 31                 row0_has_zero = true;

 32             }

 33         }

 34         for (i = 1; i < m; ++i) {

 35             for (j = 1; j < n; ++j) {

 36                 if (board[i][j] == 0) {

 37                     board[i][0] = 0;

 38                     board[0][j] = 0;

 39                 }

 40             }

 41         }

 42         for (i = 1; i < m; ++i) {

 43             if (board[i][0] == 0) {

 44                 for (j = 1; j < n; ++j) {

 45                     board[i][j] = 0;

 46                 }

 47             }

 48         }

 49         for (j = 1; j < n; ++j) {

 50             if (board[0][j] == 0) {

 51                 for (i = 1; i < m; ++i) {

 52                     board[i][j] = 0;

 53                 }

 54             }

 55         }

 56         if (row0_has_zero) {

 57             for (j = 0; j < n; ++j) {

 58                 board[0][j] = 0;

 59             }

 60         }

 61         if (col0_has_zero) {

 62             for (i = 0; i < m; ++i) {

 63                 board[i][0] = 0;

 64             }

 65         }

 66     }

 67 };

 68 

 69 int main()

 70 {

 71     vector<vector<int> > board;

 72     int i, j;

 73     int m, n;

 74     Solution sol;

 75     

 76     while (scanf("%d%d", &m, &n) == 2 && (m || n)) {

 77         board.resize(m);

 78         for (i = 0; i < m; ++i) {

 79             board[i].resize(n);

 80         }

 81         for (i = 0; i < m; ++i) {

 82             for (j = 0; j < n; ++j) {

 83                 scanf("%d", &board[i][j]);

 84             }

 85         }

 86         sol.setMatrixZero(board);

 87         for (i = 0; i < m; ++i) {

 88             for (j = 0; j < n; ++j) {

 89                 if (j > 0) {

 90                     printf(" %d", board[i][j]);

 91                 } else {

 92                     printf("%d", board[i][j]);

 93                 }

 94             }

 95             printf("\n");

 96         }

 97         printf("\n");

 98     }

 99     

100     return 0;

101 }