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.

受mitbbs版上某楼的启发，把这道题转化为之前做过的 maximal rectangle in histogram来做。

class Solution {public:    int maximalRectangle(vector<vector<char> > &matrix) {        // Start typing your C/C++ solution below        // DO NOT write int main() function                int nSize = matrix.size();        if(nSize == 0){            return 0;        }        vector<vector<char> > heights;        heights.push_back(vector<char>(matrix[0].size(), 0));        int max = 0;        for(int i = 0; i< nSize; ++i){            vector<char> row;            for(int j = 0; j< matrix[i].size(); ++j){                if(matrix[i][j] == '1'){                    row.push_back(heights[i][j] + 1);                }                else{                    row.push_back(0);                }            }            heights.push_back(row);        }        for(int i = 1; i <= nSize; ++i){            int temp = largestRectangleArea(heights[i]);            max = max > temp? max: temp;        }        return max;    }    int largestRectangleArea(vector<char> &height) {          // Start typing your C/C++ solution below          // DO NOT write int main() function          int nSize = height.size();          int l[nSize + 1];//left bar position          int h[nSize + 1];//height of the left bar          l[0] = h[0] = 0;          int cur_h = 0;          int max = 0;          int top = 0;          for(int i = 0 ; i<= nSize; ++i){              cur_h = i==nSize? 0: height[i];              int j = i;              for(; h[top] > cur_h; j = l[top--]){                  int temp = (i - l[top]) * h[top];                  if(temp > max){                      max = temp;                  }              }              if(cur_h > h[top]){                  h[++top] = cur_h;                  l[top] = j;              }          }          return max;      }  };

mitbbs上看到的另一个解法。

《浅谈用极大化思想解决最大子矩形问题》这篇文章 里面介绍了一种悬线法的 DP 时间复杂度是 O(mn) 的 原文图文讲解很好 实现起来也比 histogram 那个容易一些。

class Solution {public:    int maximalRectangle(vector<vector<char> > &matrix) {        if (matrix.empty()) {            return 0;        }        int n = matrix[0].size();        vector<int> H(n);        vector<int> L(n);        vector<int> R(n, n);        int ret = 0;        for (int i = 0; i < matrix.size(); ++i) {            int left = 0, right = n;            for (int j = 0; j < n; ++j) {                if (matrix[i][j] == '1') {                    ++H[j];                    L[j] = max(L[j], left);                }                else {                    left = j+1;                    H[j] = 0; L[j] = 0; R[j] = n;                }            }            for (int j = n-1; j >= 0; --j) {                if (matrix[i][j] == '1') {                    R[j] = min(R[j], right);                    ret = max(ret, H[j]*(R[j]-L[j]));                }                else {                    right = j;                }            }        }        return ret;    }};