Careercup - Facebook面试题 - 6299074475065344

2014-05-01 01:00

题目链接

原题：

Given a matrix with 's and 0's, a rectangle can be made with 's. What is the maximum area of the rectangle. 
 
 In this test case the result needs to be . 
 
How: 
 
If you see above the 's are used from the first two columns and last four rows making the area or count of 1's to be .

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 == ) {
             return ;
         }
         m = (int)v[].size();
         if (m == ) {
             return ;
         }
 
         int i, j;
         int res, max_res;
 
         histogram.resize(m);
         left.resize(m);
         right.resize(m);
         fill_n(histogram.begin(), m, );
         max_res = ;
         for (i = ; i < n; ++i) {
             for (j = ; j < m; ++j) {
                 histogram[j] = v[i][j] ? histogram[j] + v[i][j]: ;
                 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[] = ;
         for (i = ; i <= n - ; ++i) {
             j = i - ;
             left[i] = i;
             while (j >=  && histogram[i] <= histogram[j]) {
                 left[i] = left[j];
                 j = left[j] - ;
             }
         }
 
         right[n - ] = n - ;
         for (i = n - ; i >= ; --i) {
             j = i + ;
             right[i] = i;
             while (j <= n -  && histogram[i] <= histogram[j]) {
                 right[i] = right[j];
                 j = right[j] + ;
             }
         }
 
         int max_res, res;
         max_res = ;
         for (i = ; i < n; ++i) {
             res = histogram[i] * (right[i] - left[i] + );
             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) ==  && (n >  && m > )) {
         v.resize(n);
         for (i = ; i < n; ++i) {
             v[i].resize(m);
         }
         for (i = ; i < n; ++i) {
             for (j = ; j < m; ++j) {
                 scanf("%d", &v[i][j]);
             }
         }
         printf("%d\n", sol.maxRectangleWithAllOnes(v));
 
         for (i = ; i < n; ++i) {
             v[i].clear();
         }
         v.clear();
     }
 
     return ;
 }