《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 == 0) {
            return;
        }
        n = (int)board[0].size();
        if (n == 0) {
            return;
        }
        
        int i, j;
        bool row0_has_zero = false;
        bool col0_has_zero = false;
        for (i = 0; i < m; ++i) {
            if (board[i][0] == 0) {
                col0_has_zero = true;
                break;
            }
        }
        for (j = 0; j < n; ++j) {
            if (board[0][j] == 0) {
                row0_has_zero = true;
            }
        }
        for (i = 1; i < m; ++i) {
            for (j = 1; j < n; ++j) {
                if (board[i][j] == 0) {
                    board[i][0] = 0;
                    board[0][j] = 0;
                }
            }
        }
        for (i = 1; i < m; ++i) {
            if (board[i][0] == 0) {
                for (j = 1; j < n; ++j) {
                    board[i][j] = 0;
                }
            }
        }
        for (j = 1; j < n; ++j) {
            if (board[0][j] == 0) {
                for (i = 1; i < m; ++i) {
                    board[i][j] = 0;
                }
            }
        }
        if (row0_has_zero) {
            for (j = 0; j < n; ++j) {
                board[0][j] = 0;
            }
        }
        if (col0_has_zero) {
            for (i = 0; i < m; ++i) {
                board[i][0] = 0;
            }
        }
    }
};

int main()
{
    vector<vector<int> > board;
    int i, j;
    int m, n;
    Solution sol;
    
    while (scanf("%d%d", &m, &n) == 2 && (m || n)) {
        board.resize(m);
        for (i = 0; i < m; ++i) {
            board[i].resize(n);
        }
        for (i = 0; i < m; ++i) {
            for (j = 0; j < n; ++j) {
                scanf("%d", &board[i][j]);
            }
        }
        sol.setMatrixZero(board);
        for (i = 0; i < m; ++i) {
            for (j = 0; j < n; ++j) {
                if (j > 0) {
                    printf(" %d", board[i][j]);
                } else {
                    printf("%d", board[i][j]);
                }
            }
            printf("\n");
        }
        printf("\n");
    }
    
    return 0;
}