leetcode -- Maximal Rectangle TODO O(N)

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

 

[解题思路]

1.brute force

枚举所有sub-matrix(O(N^2), N = m*n) ，检查每个子矩阵是不是都是1，如果是更新最大面积，检查子矩阵是否都是1需要

花费O(N). 故总的时间为O(N^3) N = m*n

可以过小数据，大数据直接TLE

 1 public int maximalRectangle(char[][] matrix) {
 2         // Start typing your Java solution below
 3         // DO NOT write main() function
 4         int m = matrix.length;
 5         if(m == 0){
 6             return m;
 7         }
 8         int n = matrix[0].length;
 9         if(n == 0){
10             return n;
11         }
12         
13         return generateMaxArea(matrix);
14     }
15     
16     private static int generateMaxArea(char[][] matrix) {
17         int m = matrix.length;
18         int n = matrix[0].length;
19         int maxArea = 0;
20         for (int i = 1; i <= m; i++) {
21             for (int j = 1; j <= n; j++) {
22                 int subMatrixArea = enumerateSubMatrix(matrix, i, j);
23                 if (subMatrixArea > maxArea) {
24                     maxArea = subMatrixArea;
25                 }
26             }
27         }
28         return maxArea;
29     }
30 
31     public static int enumerateSubMatrix(char[][] matrix, int i, int j) {
32         int m = matrix.length;
33         int n = matrix[0].length;
34         int subMatrixArea = 0;
35         for (int p = 0; p <= (m - i); p++) {
36             for (int q = 0; q <= (n - j); q++) {
37                 int area = getSubMatrixArea(matrix, p, q, p + i - 1, q + j - 1);
38                 if (area > subMatrixArea) {
39                     subMatrixArea = area;
40                 }
41             }
42         }
43         return subMatrixArea;
44     }
45 
46     private static int getSubMatrixArea(char[][] matrix, int p, int q, int i,
47             int j) {
48         for (int m = p; m <= i; m++) {
49             for (int n = q; n <= j; n++) {
50                 if (matrix[m][n] == '0') {
51                     return 0;
52                 }
53             }
54         }
55 
56         return (i - p + 1) * (j - q + 1);
57     }

 2.DP

令dp[i][j]表示点(i,j)开始向右连续1的个数，花费O(M*N)的时间可以计算出来

接着从每个点开始，将该点作为矩形左上角点，从该点开始向下扫描直到最后一行或者dp[k][j] == 0

每次计算一个矩形的面积，与最大面积进行比较，如最大面积小于当前面积则进行更新，总的时间复杂度为O(M*N*M)

 1 public int maximalRectangle(char[][] matrix) {
 2         // Start typing your Java solution below
 3         // DO NOT write main() function
 4         int m = matrix.length;
 5         if(m == 0){
 6             return m;
 7         }
 8         int n = matrix[0].length;
 9         int[][] dp = new int[m][n];
10         
11         for(int i = 0; i < m; i++){
12             for(int j = 0; j < n; j++){
13                 if(matrix[i][j] == '0'){
14                     continue;
15                 } else {
16                     dp[i][j] = 1;
17                     int k = j + 1;
18                     while(k < n && (matrix[i][k] == '1')){
19                         dp[i][j] += 1;
20                         k++;
21                     }
22                 }
23             }
24         }
25         int maxArea = 0;
26         for(int i = 0; i < m; i++){
27             for(int j = 0; j < n; j++){
28                 if(dp[i][j] == 0){
29                     continue;
30                 } else{
31                     int area = 0, minDpCol = dp[i][j];
32                     for(int k = i; k < m && dp[k][j] > 0; k++){
33                         if(dp[k][j] < minDpCol){
34                             minDpCol = dp[k][j];
35                         }
36                         area = (k - i + 1) * minDpCol;
37                         if(area > maxArea){
38                             maxArea = area;
39                         }
40                     }
41                 }
42             }
43         }
44         return maxArea;
45     }