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

 public int maximalRectangle(char[][] matrix) {
         // Start typing your Java solution below
         // DO NOT write main() function
         int m = matrix.length;
         if(m == 0){
             return m;
         }
         int n = matrix[0].length;
         if(n == 0){
             return n;
         }

         return generateMaxArea(matrix);
     }

     private static int generateMaxArea(char[][] matrix) {
         int m = matrix.length;
         int n = matrix[0].length;
         int maxArea = 0;
         for (int i = 1; i <= m; i++) {
             for (int j = 1; j <= n; j++) {
                 int subMatrixArea = enumerateSubMatrix(matrix, i, j);
                 if (subMatrixArea > maxArea) {
                     maxArea = subMatrixArea;
                 }
             }
         }
         return maxArea;
     }

     public static int enumerateSubMatrix(char[][] matrix, int i, int j) {
         int m = matrix.length;
         int n = matrix[0].length;
         int subMatrixArea = 0;
         for (int p = 0; p <= (m - i); p++) {
             for (int q = 0; q <= (n - j); q++) {
                 int area = getSubMatrixArea(matrix, p, q, p + i - 1, q + j - 1);
                 if (area > subMatrixArea) {
                     subMatrixArea = area;
                 }
             }
         }
         return subMatrixArea;
     }

     private static int getSubMatrixArea(char[][] matrix, int p, int q, int i,
             int j) {
         for (int m = p; m <= i; m++) {
             for (int n = q; n <= j; n++) {
                 if (matrix[m][n] == '0') {
                     return 0;
                 }
             }
         }

         return (i - p + 1) * (j - q + 1);
     }

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)

 public int maximalRectangle(char[][] matrix) {
         // Start typing your Java solution below
         // DO NOT write main() function
         int m = matrix.length;
         if(m == 0){
             return m;
         }
         int n = matrix[0].length;
         int[][] dp = new int[m][n];

         for(int i = 0; i < m; i++){
             for(int j = 0; j < n; j++){
                 if(matrix[i][j] == '0'){
                     continue;
                 } else {
                     dp[i][j] = 1;
                     int k = j + 1;
                     while(k < n && (matrix[i][k] == '1')){
                         dp[i][j] += 1;
                         k++;
                     }
                 }
             }
         }
         int maxArea = 0;
         for(int i = 0; i < m; i++){
             for(int j = 0; j < n; j++){
                 if(dp[i][j] == 0){
                     continue;
                 } else{
                     int area = 0, minDpCol = dp[i][j];
                     for(int k = i; k < m && dp[k][j] > 0; k++){
                         if(dp[k][j] < minDpCol){
                             minDpCol = dp[k][j];
                         }
                         area = (k - i + 1) * minDpCol;
                         if(area > maxArea){
                             maxArea = area;
                         }
                     }
                 }
             }
         }
         return maxArea;
     }