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 };

 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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posted @ 2013-04-26 12:23 ying_vincent 阅读(297) 评论(0) 收藏举报

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