LeetCode: Maximal Rectangle

Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area.

O(n^3)方法，348 milli secs

 

class Solution {public:    int maximalRectangle(vector<vector<char> > &matrix) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int y = matrix.size();        if (y == 0) return 0;        int x = matrix[0].size();        if (x == 0) return 0;                vector< vector<int> > result(y, vector<int>(x));        for (int j = 0; j < y; ++j)        {            for (int i = 0; i < x; ++i)            {                result[j][i] = '0' == matrix[j][i] ? 0 : 1;            }        }                for (int j = 0; j < y; ++j)        {            for (int i = 1; i < x; ++i)            {                result[j][i] += (0 == result[j][i] ? 0 : result[j][i-1]);            }        }                int max = 0;        for (int i = 0; i < x; ++i)        {            for (int j = 0; j < y; ++j)            {                int maxWidth = result[j][i];                  for (int k = j; k < y; ++k)                {                    maxWidth = maxWidth <  result[k][i] ? maxWidth : result[k][i];                    max = max > maxWidth*(k-j+1) ? max : maxWidth*(k-j+1);                }            }        }        return max;    }};

O(n^2)方法： 44 milli secs

class Solution {public:    int maximalRectangle(vector<vector<char> > &matrix) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int y = matrix.size();        if (y == 0) return 0;        int x = matrix[0].size();        if (x == 0) return 0;                vector< vector<int> > result(y, vector<int>(x));        for (int j = 0; j < y; ++j)        {            int tmp = 0;            for (int i = 0; i < x; ++i)            {                if ('0' == matrix[j][i])                    tmp =  0;                else                    ++tmp;                result[j][i] = tmp;            }        }                int max = 0;        int* l = new int[y];          int* r = new int[y];          for (int i = 0; i < x; ++i)        {            l[0] = 0;            for (int j = 1; j < y; ++j)            {                int k = j;                  while(k > 0 && result[j][i] <= result[k-1][i])                      k = l[k-1];                  l[j] = k;              }                        r[y-1] = y - 1;             for(int j = y-2; j >= 0; --j)              {                  int k = j;                  while(k < y - 1 && result[j][i] <= result[k+1][i])                      k = r[k+1];                  r[j] = k;              }                         for (int j = 0; j < y; ++j)              {                  int ar = (r[j] - l[j] + 1) * result[j][i];                  max = max > ar ? max : ar;                 }          }        return max;    }};