• [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.

    思路:

    动态规划:

    f[i][j] 表示在第i行中以(i,j)点结束点的前面有多少个连续的1,就是用f[i][j]来记录i行以j列为结尾,往前连续的1的个数。

    然后用O(n^3)的循环来找以(i, j)为右下角的矩形最大的矩形的面积。

    例如数组为

    0  0  1

    1  1  1

    0  1  1

    那么f数组为

    0  0  1

    1  2  3

    0  1  2

    在计算f[i][j]时使用的是一维动态规划

    计算(i, j)为右下角的矩形最大的矩形的面积时,就是就去第0行到i行中以j结尾矩形的最大值,就是之前求过的 Largest Rectangle in Histogram

    相当于是每一列,求一次Largest Rectangle in Histogram,然后求其所有列的最大值。。

    方法一: O(n^3)

    转载 http://www.cnblogs.com/remlostime/archive/2012/11/12/2766566.html

     1 class Solution {
     2     private:
     3         int f[1000][1000];
     4     public:
     5         int maximalRectangle(vector<vector<char> > &matrix) {
     6             // Start typing your C/C++ solution below
     7             // DO NOT write int main() function
     8             for(int i = 0; i < matrix.size(); i++)
     9                 f[i][0] = matrix[i][0] == '1' ? 1 : 0;
    10 
    11             for(int i = 0; i < matrix.size(); i++)
    12                 for(int j = 1; j < matrix[i].size(); j++)
    13                     f[i][j] = matrix[i][j] == '1' ? f[i][j-1] + 1 : 0;
    14 
    15             int ret = 0;
    16             for(int i = 0; i < matrix.size(); i++)
    17             {
    18                 for(int j = 0; j < matrix[i].size(); j++)
    19                 {
    20                     int wid = INT_MAX;
    21                     for(int k = i; k >=0; k--)
    22                     {
    23                         if(f[i][j] == 0)
    24                             break;
    25                         wid = min(wid, f[k][j]);
    26                         ret = max(ret, wid* (i-k +1));
    27                     }
    28 
    29                 }
    30             }
    31             return ret;
    32 
    33         }
    34 };

     方法二,基于largestRectangleArea

    将matrxi看出n行,求出每行的高度,然后调用largestRectangleArea

    求解每行的高度的时候,使用的动态规划方法和方法一类似,

    f[i][j]表示第i列中以(i,j)为结束连续1的个数

    重点是

    if(matrix[i][j] == '1')
    f[i][j] = f[i-1][j] + 1;
    else
    f[i][j] = 0;

     共n行,每行调用largestRectangleArea的复杂度为O(n),所以是O(n^2),空间复杂度由于有矩阵f的存在,也是O(n^2)

     1 class Solution
     2 {
     3     public:
     4         int largestRectangleArea(vector<int> &height) {
     5 
     6             //add 0 to the end
     7             height.push_back(0);
     8 
     9             int size = height.size(); //this new_size is 1 + old_size
    10             stack<int> st;
    11 
    12             int max_area = 0;
    13 
    14 
    15 
    16             for(int i = 0; i< size; )
    17             {
    18                 if(st.empty() || height[i] > height[st.top()])
    19                 {
    20                     st.push(i);
    21                     i++;
    22                 }
    23                 else
    24                 {
    25                     //st must be not empty here
    26                     // i can't ++; handle many times perhaps
    27 
    28                     int idx = st.top();
    29                     st.pop();
    30 
    31                     int area;
    32 
    33                     if(st.empty())
    34                     {
    35                         area = height[idx] * i;
    36                     }
    37                     else
    38                     {
    39                         area = height[idx] * (i-st.top()-1);
    40                     }
    41 
    42                     max_area = max(max_area, area);
    43 
    44                 }
    45             }
    46             return max_area;
    47         }
    48 
    49         int maximalRectangle(vector<vector<char> > &matrix)
    50         {
    51             if(matrix.empty() || matrix[0].empty())
    52                 return 0;
    53             size_t row = matrix.size();
    54             size_t col = matrix[0].size();
    55 
    56             //construct matrix f;
    57             vector<vector<int> > f(row);
    58             vector<int> array;
    59             array.resize(col, 0);
    60 
    61             for(int i = 0; i < row; i++)
    62                 f[i] = array;
    63 
    64             // initial f[0], the first row
    65             for(int j = 0; j < col; j++)
    66             {
    67                 if(matrix[0][j] == '1')
    68                     f[0][j] = 1;
    69                 else
    70                     f[0][j] = 0;
    71             }
    72 
    73             for(int i = 1; i < row; i++)
    74             {
    75                 for(int j = 0; j < col; j++)
    76                 {
    77                     if(matrix[i][j] == '1')
    78                         f[i][j] = f[i-1][j] + 1;
    79                     else
    80                         f[i][j] = 0;
    81                 }
    82             }
    83 
    84             int ret = 0;
    85             for(int i = 0; i < row; i++)
    86             {
    87 
    88                 ret = max(ret, largestRectangleArea(f[i]));
    89             }
    90 
    91             return ret;
    92 
    93         }
    94 } ;

     适当的优化:

      由于f[i][j] = f[i-1][j] + 1;的关系,二期在计算第i行时,第i-1行的数据有用,再之前的数据就没有了,所以我么可以不用保留整个col*row的矩阵,而只用一个数组就够了

      空间复杂度O(n),时间复杂度还是O(n^2)

     1 class Solution
     2 {
     3     public:
     4         int largestRectangleArea(vector<int> &height) {
     5 
     6             //add 0 to the end
     7             height.push_back(0);
     8 
     9             int size = height.size(); //this new_size is 1 + old_size
    10             stack<int> st;
    11 
    12             int max_area = 0;
    13 
    14 
    15 
    16             for(int i = 0; i< size; )
    17             {
    18                 if(st.empty() || height[i] > height[st.top()])
    19                 {
    20                     st.push(i);
    21                     i++;
    22                 }
    23                 else
    24                 {
    25                     //st must be not empty here
    26                     // i can't ++; handle many times perhaps
    27 
    28                     int idx = st.top();
    29                     st.pop();
    30 
    31                     int area;
    32 
    33                     if(st.empty())
    34                     {
    35                         area = height[idx] * i;
    36                     }
    37                     else
    38                     {
    39                         area = height[idx] * (i-st.top()-1);
    40                     }
    41 
    42                     max_area = max(max_area, area);
    43 
    44                 }
    45             }
    46             return max_area;
    47         }
    48 
    49         int maximalRectangle(vector<vector<char> > &matrix)
    50         {
    51             if(matrix.empty() || matrix[0].empty())
    52                 return 0;
    53             size_t row = matrix.size();
    54             size_t col = matrix[0].size();
    55 
    56             vector<int> f;
    57             f.resize(col, 0);
    58 
    59             int ret = 0;
    60             for(int i = 0; i < row; i++)
    61             {
    62                 // update f
    63                 for(int j = 0; j < col; j++)
    64                 {
    65                     // the first row 
    66                     if(i == 0)
    67                     {
    68                         if(matrix[0][j] == '1')
    69                             f[j] = 1;
    70                         else
    71                             f[j] = 0;
    72                     }
    73                     // other rows
    74                     else
    75                     {
    76                         if(matrix[i][j] == '1')
    77                             f[j] = f[j] + 1;
    78                         else
    79                             f[j] = 0;
    80                     }
    81                 }
    82 
    83                 ret = max(ret, largestRectangleArea(f));
    84             }
    85 
    86             return ret;
    87 
    88         }
    89 } ;
  • 相关阅读:
    PHP入门03 -- 数组与数据结构
    PHP入门02 -- 函数
    PHP入门01 -- 基本语法
    node文章
    Mongodb08
    Mongodb07
    ISO处理jq事件
    map
    Django自定义模板
    JS this指向
  • 原文地址:https://www.cnblogs.com/diegodu/p/3823868.html
Copyright © 2020-2023  润新知