LeetCode: Maximal Rectangle

这道题把二维图转换成行数个的柱状图题，就简单了

 1 class Solution {

 2 public:

 3     int maximalRectangle(vector<vector<char> > &matrix) {

 4         int x = matrix.size();

 5         if(0 == x)    return 0;

 6         int y = matrix[0].size();

 7         if(0 == y)    return 0;

 8         vector<vector<int> > result(x, vector<int>(y));

 9         for(int i = 0; i < x; ++i)

10             for(int j = 0; j < y; ++j)

11                 result[i][j] = '0' == matrix[i][j]? 0: 1;

12         for(int i = 1; i < x; ++i)

13             for(int j = 0; j < y; ++j)

14                 result[i][j] += 0 == result[i][j]? 0: result[i-1][j];

15         int ret = 0;

16         for(int i = 0; i < x; ++i)

17             ret = max(ret, maxArea(result[i]));

18         return ret;

19     }

20 

21     int maxArea(vector<int>& line)

22     {

23         stack<int> S;

24         line.push_back(0);

25         int sum = 0;

26         for (int i = 0; i < line.size(); i++) {

27             if (S.empty() || line[i] > line[S.top()]) S.push(i);

28             else {

29                 int tmp = S.top();

30                 S.pop();

31                 sum = max(sum, line[tmp]*(S.empty()? i : i-S.top()-1));

32                 i--;

33             }

34         }

35         return sum;

36     }

37 };

 C#

 1 public class Solution {

 2     public int MaximalRectangle(char[,] matrix) {

 3         int m = matrix.GetLength(0);

 4         int n = matrix.GetLength(1);

 5         if (m == 0 || n == 0) return 0;

 6         int[,] result = new int[m, n];

 7         for (int i = 0; i < m; i++) {

 8             for (int j = 0; j < n; j++) {

 9                 result[i, j] = '0' == matrix[i, j]? 0 : 1;

10             }

11         }

12         for (int i = 1; i < m; i++) {

13             for (int j = 0; j < n; j++) {

14                 result[i, j] += 0 == result[i, j]? 0 : result[i-1, j];

15             }

16         }

17         int ans = 0;

18         for (int i = 0; i < m; i++) {

19             ans = Math.Max(ans, maxArea(result, i));

20         }

21         return ans;

22     }

23     public int maxArea(int[,] result, int line) {

24         Stack<int> S = new Stack<int>();

25         List<int> lineCopy = new List<int>();

26         for (int i = 0; i < result.GetLength(1); i++) lineCopy.Add(result[line, i]);

27         lineCopy.Add(0);

28         int sum = 0;

29         for (int i = 0; i < lineCopy.Count; i++) {

30             if (S.Count == 0 || lineCopy[i] > lineCopy[S.Peek()]) S.Push(i);

31             else {

32                 int peek = S.Peek();

33                 S.Pop();

34                 sum = Math.Max(sum, lineCopy[peek] * (S.Count == 0? i : i - S.Peek() - 1));

35                 i--;

36             }

37         }

38         return sum;

39     }

40 }

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