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;
    }