85. Maximal Rectangle

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

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1. O(n^3) 

2. 很想84， largest rectangle in histogra,

每一行存以当前行为底的柱状图， 求最大矩形面积。

O(N^2)

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10/01 YC version

public class Solution {
    public int maximalRectangle(char[][] matrix) {
        
        int rowN = matrix.length;
        int calN = rowN==0 ? 0 : matrix[0].length; // check if empty matrix
        
        int[][] helperMatrix = new int[rowN][calN];
        
        for(int i=0; i<rowN; i++){
            for(int j=0; j<calN; j++){
                
                if(i == 0){
                    if(matrix[i][j] == '1') helperMatrix[i][j] = 1;
                }else{
                    if(matrix[i][j] == '1')
                        helperMatrix[i][j] = helperMatrix[i-1][j] + 1;
                    else
                        helperMatrix[i][j] = 0;
                }
                
            }
        }
        
        int maxArea = 0;
        
        for(int i=0; i<rowN; i++){
            int area = largestRectangleArea(helperMatrix, i);
            // update
            if(area > maxArea)  maxArea = area;
        }
        
        return maxArea;
        
    }
    
    
    
    public int largestRectangleArea(int[][] matrix, int row) {
        
        int maxArea = 0;
        Stack<Integer> stack = new Stack<Integer>();
        
        int i=0;
        while(i<matrix[row].length){
            
            if(stack.isEmpty() || matrix[row][stack.peek()] <=  matrix[row][i]){
                stack.push(i);
                i++;
            }else{
                // current min height index t
                int t = stack.pop();
                int area = (stack.isEmpty() ? i : i-stack.peek()-1) *  matrix[row][t];
                // update
                if(area > maxArea)  maxArea = area;
            }
        }
        
        while(!stack.isEmpty()){
            int t = stack.pop();
            int area = (stack.isEmpty() ? i : i-stack.peek()-1) *  matrix[row][t];
            // update
            if(area > maxArea)  maxArea = area;
        }
        
        
        return maxArea;
    }
}

 

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public class Solution {
    public int maximalRectangle(char[][] matrix) {
        int m = matrix.length;
        int n = m == 0 ? 0 : matrix[0].length;
        int[][] height = new int[m][n + 1];
　　　　　//　（有人说！）
        //actually we know that height can just be a int[n+1], 
        //however, in that case, we have to write the 2 parts together in row traverse,
        //which, leetcode just doesn't make you pass big set
        //leetcode是喜欢分开写循环的，即使时间复杂度一样，即使可以节约空间
        int maxArea = 0;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++) {
                if(matrix[i][j] == '0'){
                    height[i][j] = 0;
                }else {
                    height[i][j] = i == 0 ? 1 : height[i - 1][j] + 1;
                }
            }
        }
        for(int i = 0; i < m; i++){
            int area = maxAreaInHist(height[i]);
            if(area > maxArea){
                maxArea = area;
            }
        }
        return maxArea;
     }
     
     private int maxAreaInHist(int[] height){
         Stack<Integer> stack = new Stack<Integer>();
         int i = 0;
         int maxArea = 0;
         while(i < height.length){
             if(stack.isEmpty() || height[stack.peek()] <= height[i]){
                 stack.push(i++);
             }else {
                 int t = stack.pop();
                 maxArea = Math.max(maxArea, height[t] * (stack.isEmpty() ? i : i - stack.peek() - 1));
             }
         }
         return maxArea;
     }
}

 

转载于:https://www.cnblogs.com/ycled/p/3330156.html