Question

85. 最大矩形

English Version

题目描述

给定一个仅包含 0 和 1 、大小为 rows x cols 的二维二进制矩阵，找出只包含 1 的最大矩形，并返回其面积。

 

示例 1：

输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
输出：6
解释：最大矩形如上图所示。

示例 2：

输入：matrix = [["0"]]
输出：0

示例 3：

输入：matrix = [["1"]]
输出：1

 

提示：

• rows == matrix.length
• cols == matrix[0].length
• 1 <= row, cols <= 200
• matrix[i][j] 为 '0' 或 '1'

解法

方法一：单调栈

我们把每一行视为柱状图的底部，对每一行求柱状图的最大面积即可。

时间复杂度 $O(m \times n)$，其中 $m$ 表示 $matrix$ 的行数，$n$ 表示 $matrix$ 的列数。

class Solution:
  def maximalRectangle(self, matrix: List[List[str]]) -> int:
    heights = [0] * len(matrix[0])
    ans = 0
    for row in matrix:
      for j, v in enumerate(row):
        if v == "1":
          heights[j] += 1
        else:
          heights[j] = 0
      ans = max(ans, self.largestRectangleArea(heights))
    return ans

  def largestRectangleArea(self, heights: List[int]) -> int:
    n = len(heights)
    stk = []
    left = [-1] * n
    right = [n] * n
    for i, h in enumerate(heights):
      while stk and heights[stk[-1]] >= h:
        stk.pop()
      if stk:
        left[i] = stk[-1]
      stk.append(i)
    stk = []
    for i in range(n - 1, -1, -1):
      h = heights[i]
      while stk and heights[stk[-1]] >= h:
        stk.pop()
      if stk:
        right[i] = stk[-1]
      stk.append(i)
    return max(h * (right[i] - left[i] - 1) for i, h in enumerate(heights))

class Solution {
  public int maximalRectangle(char[][] matrix) {
    int n = matrix[0].length;
    int[] heights = new int[n];
    int ans = 0;
    for (var row : matrix) {
      for (int j = 0; j < n; ++j) {
        if (row[j] == '1') {
          heights[j] += 1;
        } else {
          heights[j] = 0;
        }
      }
      ans = Math.max(ans, largestRectangleArea(heights));
    }
    return ans;
  }

  private int largestRectangleArea(int[] heights) {
    int res = 0, n = heights.length;
    Deque<Integer> stk = new ArrayDeque<>();
    int[] left = new int[n];
    int[] right = new int[n];
    Arrays.fill(right, n);
    for (int i = 0; i < n; ++i) {
      while (!stk.isEmpty() && heights[stk.peek()] >= heights[i]) {
        right[stk.pop()] = i;
      }
      left[i] = stk.isEmpty() ? -1 : stk.peek();
      stk.push(i);
    }
    for (int i = 0; i < n; ++i) {
      res = Math.max(res, heights[i] * (right[i] - left[i] - 1));
    }
    return res;
  }
}

class Solution {
public:
  int maximalRectangle(vector<vector<char>>& matrix) {
    int n = matrix[0].size();
    vector<int> heights(n);
    int ans = 0;
    for (auto& row : matrix) {
      for (int j = 0; j < n; ++j) {
        if (row[j] == '1')
          ++heights[j];
        else
          heights[j] = 0;
      }
      ans = max(ans, largestRectangleArea(heights));
    }
    return ans;
  }

  int largestRectangleArea(vector<int>& heights) {
    int res = 0, n = heights.size();
    stack<int> stk;
    vector<int> left(n, -1);
    vector<int> right(n, n);
    for (int i = 0; i < n; ++i) {
      while (!stk.empty() && heights[stk.top()] >= heights[i]) {
        right[stk.top()] = i;
        stk.pop();
      }
      if (!stk.empty()) left[i] = stk.top();
      stk.push(i);
    }
    for (int i = 0; i < n; ++i)
      res = max(res, heights[i] * (right[i] - left[i] - 1));
    return res;
  }
};

func maximalRectangle(matrix [][]byte) int {
  n := len(matrix[0])
  heights := make([]int, n)
  ans := 0
  for _, row := range matrix {
    for j, v := range row {
      if v == '1' {
        heights[j]++
      } else {
        heights[j] = 0
      }
    }
    ans = max(ans, largestRectangleArea(heights))
  }
  return ans
}

func largestRectangleArea(heights []int) int {
  res, n := 0, len(heights)
  var stk []int
  left, right := make([]int, n), make([]int, n)
  for i := range right {
    right[i] = n
  }
  for i, h := range heights {
    for len(stk) > 0 && heights[stk[len(stk)-1]] >= h {
      right[stk[len(stk)-1]] = i
      stk = stk[:len(stk)-1]
    }
    if len(stk) > 0 {
      left[i] = stk[len(stk)-1]
    } else {
      left[i] = -1
    }
    stk = append(stk, i)
  }
  for i, h := range heights {
    res = max(res, h*(right[i]-left[i]-1))
  }
  return res
}

impl Solution {
  #[allow(dead_code)]
  pub fn maximal_rectangle(matrix: Vec<Vec<char>>) -> i32 {
    let n = matrix[0].len();
    let mut heights = vec![0; n];
    let mut ret = -1;

    for row in &matrix {
      Self::array_builder(row, &mut heights);
      ret = std::cmp::max(ret, Self::largest_rectangle_area(heights.clone()));
    }

    ret
  }

  /// Helper function, build the heights array according to the input
  #[allow(dead_code)]
  fn array_builder(input: &Vec<char>, heights: &mut Vec<i32>) {
    for (i, &c) in input.iter().enumerate() {
      heights[i] += match c {
        '1' => 1,
        '0' => {
          heights[i] = 0;
          0
        }
        _ => panic!("This is impossible"),
      };
    }
  }

  /// Helper function, see: https://leetcode.com/problems/largest-rectangle-in-histogram/ for details
  #[allow(dead_code)]
  fn largest_rectangle_area(heights: Vec<i32>) -> i32 {
    let n = heights.len();
    let mut left = vec![-1; n];
    let mut right = vec![-1; n];
    let mut stack: Vec<(usize, i32)> = Vec::new();
    let mut ret = -1;

    // Build left vector
    for (i, h) in heights.iter().enumerate() {
      while !stack.is_empty() && stack.last().unwrap().1 >= *h {
        stack.pop();
      }
      if stack.is_empty() {
        left[i] = -1;
      } else {
        left[i] = stack.last().unwrap().0 as i32;
      }
      stack.push((i, *h));
    }

    stack.clear();

    // Build right vector
    for (i, h) in heights.iter().enumerate().rev() {
      while !stack.is_empty() && stack.last().unwrap().1 >= *h {
        stack.pop();
      }
      if stack.is_empty() {
        right[i] = n as i32;
      } else {
        right[i] = stack.last().unwrap().0 as i32;
      }
      stack.push((i, *h));
    }

    // Calculate the max area
    for (i, h) in heights.iter().enumerate() {
      ret = std::cmp::max(ret, (right[i] - left[i] - 1) * *h);
    }

    ret
  }
}

using System;
using System.Collections.Generic;
using System.Linq;

public class Solution {
  private int MaximalRectangleHistagram(int[] height) {
    var stack = new Stack<int>();
    var result = 0;
    var i = 0;
    while (i < height.Length || stack.Any())
    {
      if (!stack.Any() || (i < height.Length && height[stack.Peek()] < height[i]))
      {
        stack.Push(i);
        ++i;
      }
      else
      {
        var previousIndex = stack.Pop();
        var area = height[previousIndex] * (stack.Any() ? (i - stack.Peek() - 1) : i);
        result = Math.Max(result, area);
      }
    }

    return result;
  }

  public int MaximalRectangle(char[][] matrix) {
    var lenI = matrix.Length;
    var lenJ = lenI == 0 ? 0 : matrix[0].Length;
    var height = new int[lenJ];
    var result = 0;
    for (var i = 0; i < lenI; ++i)
    {
      for (var j = 0; j < lenJ; ++j)
      {
        if (matrix[i][j] == '1')
        {
          ++height[j];
        }
        else
        {
          height[j] = 0;
        }
      }
      result = Math.Max(result, MaximalRectangleHistagram(height));
    }
    return result;
  }
}