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：

  1 // http://www.careercup.com/question?id=6299074475065344

  2 #include <vector>

  3 using namespace std;

  4 

  5 class Solution {

  6 public:

  7     int maxRectangleWithAllOnes(vector<vector<int> > &v) {

  8         n = (int)v.size();

  9         if (n == 0) {

 10             return 0;

 11         }

 12         m = (int)v[0].size();

 13         if (m == 0) {

 14             return 0;

 15         }

 16         

 17         int i, j;

 18         int res, max_res;

 19         

 20         histogram.resize(m);

 21         left.resize(m);

 22         right.resize(m);

 23         fill_n(histogram.begin(), m, 0);

 24         max_res = 0;

 25         for (i = 0; i < n; ++i) {

 26             for (j = 0; j < m; ++j) {

 27                 histogram[j] = v[i][j] ? histogram[j] + v[i][j]: 0;

 28                 res = maxRectangleInHistogram(histogram);

 29                 max_res = res > max_res ? res : max_res;

 30             }

 31         }

 32         

 33         histogram.clear();

 34         left.clear();

 35         right.clear();

 36         

 37         return max_res;

 38     };

 39 private:

 40     vector<int> histogram;

 41     vector<int> left;

 42     vector<int> right;

 43     int n, m;

 44     

 45     int maxRectangleInHistogram(vector<int> &histogram) {

 46         int i;

 47         int j;

 48         

 49         left[0] = 0;

 50         for (i = 1; i <= n - 1; ++i) {

 51             j = i - 1;

 52             left[i] = i;

 53             while (j >= 0 && histogram[i] <= histogram[j]) {

 54                 left[i] = left[j];

 55                 j = left[j] - 1;

 56             }

 57         }

 58         

 59         right[n - 1] = n - 1;

 60         for (i = n - 2; i >= 0; --i) {

 61             j = i + 1;

 62             right[i] = i;

 63             while (j <= n - 1 && histogram[i] <= histogram[j]) {

 64                 right[i] = right[j];

 65                 j = right[j] + 1;

 66             }

 67         }

 68         

 69         int max_res, res;

 70         max_res = 0;

 71         for (i = 0; i < n; ++i) {

 72             res = histogram[i] * (right[i] - left[i] + 1);

 73             max_res = res > max_res ? res : max_res;

 74         }

 75         

 76         return max_res;

 77     };

 78 };

 79 

 80 int main()

 81 {

 82     int n, m;

 83     int i, j;

 84     vector<vector<int> > v;

 85     Solution sol;

 86     

 87     while (scanf("%d%d", &n, &m) == 2 && (n > 0 && m > 0)) {

 88         v.resize(n);

 89         for (i = 0; i < n; ++i) {

 90             v[i].resize(m);

 91         }

 92         for (i = 0; i < n; ++i) {

 93             for (j = 0; j < m; ++j) {

 94                 scanf("%d", &v[i][j]);

 95             }

 96         }

 97         printf("%d\n", sol.maxRectangleWithAllOnes(v));

 98         

 99         for (i = 0; i < n; ++i) {

100             v[i].clear();

101         }

102         v.clear();

103     }

104     

105     return 0;

106 }