Careercup - Facebook面试题 - 6299074475065344

2014-05-01 01:00

题目链接

原题：

Given a matrix with 1's and 0's, a rectangle can be made with 1's. What is the maximum area of the rectangle. 

00010 
11100 
11110 
11000 
11010 In this test case the result needs to be 8. 

How: 
00010 00010 
11100 11 100 
11110 11 110 
11000 11 000 
11010 11 010 

If you see above the 11's are used from the first two columns and last four rows making the area or count of 1's to be 8.

题目：给定一个‘0’、‘1’构成的矩阵，找出其中全部由‘1’构成的子矩阵中面积最大的一个。

解法：这是Leetcode上有的题目，请参考我的题解LeetCode - Maximal Rectangle。

代码：

// http://www.careercup.com/question?id=6299074475065344
#include <vector>
using namespace std;

class Solution {
public:
    int maxRectangleWithAllOnes(vector<vector<int> > &v) {
        n = (int)v.size();
        if (n == 0) {
            return 0;
        }
        m = (int)v[0].size();
        if (m == 0) {
            return 0;
        }
        
        int i, j;
        int res, max_res;
        
        histogram.resize(m);
        left.resize(m);
        right.resize(m);
        fill_n(histogram.begin(), m, 0);
        max_res = 0;
        for (i = 0; i < n; ++i) {
            for (j = 0; j < m; ++j) {
                histogram[j] = v[i][j] ? histogram[j] + v[i][j]: 0;
                res = maxRectangleInHistogram(histogram);
                max_res = res > max_res ? res : max_res;
            }
        }
        
        histogram.clear();
        left.clear();
        right.clear();
        
        return max_res;
    };
private:
    vector<int> histogram;
    vector<int> left;
    vector<int> right;
    int n, m;
    
    int maxRectangleInHistogram(vector<int> &histogram) {
        int i;
        int j;
        
        left[0] = 0;
        for (i = 1; i <= n - 1; ++i) {
            j = i - 1;
            left[i] = i;
            while (j >= 0 && histogram[i] <= histogram[j]) {
                left[i] = left[j];
                j = left[j] - 1;
            }
        }
        
        right[n - 1] = n - 1;
        for (i = n - 2; i >= 0; --i) {
            j = i + 1;
            right[i] = i;
            while (j <= n - 1 && histogram[i] <= histogram[j]) {
                right[i] = right[j];
                j = right[j] + 1;
            }
        }
        
        int max_res, res;
        max_res = 0;
        for (i = 0; i < n; ++i) {
            res = histogram[i] * (right[i] - left[i] + 1);
            max_res = res > max_res ? res : max_res;
        }
        
        return max_res;
    };
};

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