【单调栈】【区间合并】LeetCode85:最大矩形-编程知识

作者推荐

【动态规划】【广度优先搜索】LeetCode:2617 网格图中最少访问的格子数

本文涉及的知识点

单调栈区间合并

题目

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
参数范围：
rows == matrix.length
cols == matrix[0].length
1 <= row, cols <= 200
matrix[i][j] 为 ‘0’ 或 ‘1’

分析

时间复杂度O(n²m)。枚举矩形的left和right，时间复杂度o(n^2)。指定left,right，计算连续1的数量大于等于width的行数，时间复杂度O(m)。

代码

核心代码

class Solution {
public:int maximalRectangle(vector<vector<char>>& matrix) {m_r = matrix.size();m_c = matrix.front().size();vector<vector<int>> vRightLen(m_r, vector<int>(m_c));for (int r = 0; r < m_r; r++){for (int c = m_c - 1; c >= 0; c--){if ('1' == matrix[r][c]){vRightLen[r][c] = 1 + ((m_c - 1 == c) ? 0 : vRightLen[r][c + 1]);}}}int iRet = 0;for (int left = 0; left < m_c; left++){for (int right = left; right < m_c; right++){const int width = right - left + 1;int height = 0;for (int r = 0; r < m_r; r++){if (vRightLen[r][left] < width){iRet = max(iRet, height * width);height = 0;}else{height++;}}iRet = max(iRet, height * width);}}return iRet;}int m_r, m_c;
};

测试用例

template<class T>
void Assert(const vector<T>& v1, const vector<T>& v2)
{if (v1.size() != v2.size()){assert(false);return;}for (int i = 0; i < v1.size(); i++){assert(v1[i] == v2[i]);}
}template<class T>
void Assert(const T& t1, const T& t2)
{assert(t1 == t2);
}int main()
{vector<vector<char>> matrix;int r;{Solution slu;		matrix = { {'1','0','1','0','0'},{'1','0','1','1','1'},{'1','1','1','1','1'},{'1','0','0','1','0'} };auto res = slu.maximalRectangle(matrix);Assert(6, res);}{Solution slu;matrix = { {'0'} };auto res = slu.maximalRectangle(matrix);Assert(0, res);}{Solution slu;matrix = { {'1'} };auto res = slu.maximalRectangle(matrix);Assert(1, res);}{Solution slu;matrix = { {'1','1'}};auto res = slu.maximalRectangle(matrix);Assert(2, res);}{Solution slu;matrix = { {'1'},{'1' } };auto res = slu.maximalRectangle(matrix);Assert(2, res);}
}

单调栈

枚举底部，本题就可以转化成柱形图的最大矩形

class Solution {
public:int maximalRectangle(vector<vector<char>>& matrix) {		m_c = matrix.front().size();vector<int> vHeights(m_c);int iRet = 0;for (int r = 0; r < matrix.size(); r++){for (int c = m_c - 1; c >= 0; c--){if ('1' == matrix[r][c]){vHeights[c] +=1 ;}else{vHeights[c] = 0 ;}}iRet = max(iRet, largestRectangleArea(vHeights));}return iRet;}int largestRectangleArea(vector<int>& heights) {m_c = heights.size();vector<pair<int, int>> vLeftHeightIndex;vector<int> vLeftFirstLess(m_c, -1), vRightFirstMoreEqual(m_c, m_c);//别忘记初始化for (int i = 0; i < m_c; i++){while (vLeftHeightIndex.size() && (heights[i] <= vLeftHeightIndex.back().first)){vRightFirstMoreEqual[vLeftHeightIndex.back().second] = i;vLeftHeightIndex.pop_back();}if (vLeftHeightIndex.size()){vLeftFirstLess[i] = vLeftHeightIndex.back().second;}vLeftHeightIndex.emplace_back(heights[i], i);}int iRet = 0;for (int i = 0; i < m_c; i++){iRet = max(iRet, heights[i] * (vRightFirstMoreEqual[i] - vLeftFirstLess[i] - 1));}return iRet;}int m_c;
};

2022年12月版代码

 class Solution {public:int maximalRectangle(vector<vector<char>>& matrix) {m_r = matrix.size();m_c = matrix[0].size();vector<vector<int>> leftNums;leftNums.assign(m_r, vector<int>(m_c));for (int r = 0; r < m_r; r++){for (int c = 0; c < m_c; c++){if ('0' == matrix[r][c]){leftNums[r][c] = 0;}else{leftNums[r][c] = 1 + ((c > 0) ? leftNums[r][c - 1] : 0);}}}for (int c = 0; c < m_c; c++){stack<pair<int, int>> sta;			 for (int r = 0; r < m_r; r++){int iMinR = r;while (sta.size() && (sta.top().first > leftNums[r][c])){PopStack(sta, iMinR, r);}if (sta.empty() || (sta.top().first < leftNums[r][c])){sta.emplace(leftNums[r][c], iMinR);}}while (sta.size()){int iMinR = m_r;PopStack(sta, iMinR, m_r);}}return m_iMaxArea;}void PopStack(stack<pair<int, int>>& sta,int& iMinRow,int r ){int iWidth = sta.top().first;iMinRow = sta.top().second;sta.pop();m_iMaxArea = max(m_iMaxArea, iWidth*(r - 1 - iMinRow + 1));}int m_r, m_c;int m_iMaxArea = 0;};