LeetCode开心刷题三十四天——84. Largest Rectangle in Histogram-蒲公英云

LeetCode开心刷题三十四天——84. Largest Rectangle in Histogram

Largest Rectangle in Histogram

Hard

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Given n non-negative integers representing the histogram’s bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.

Example:

Input: [2,1,5,6,2,3]
Output: 10

This problem is a bold attempt(大胆尝试) for me ,because I try multiply ways to solve this problem.I’ve been heard multi-way can make people get clear

左右遍历法：

1.O(n) 左右遍历

#include<stdio.h>
#include<iostream>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<algorithm>
#include<stack>
#include<climits>
//
using namespace std;
class Solution
{
public:
    int largestRectangleArea(vector<int>& height)
    {
        //NULL
        if(height.empty()) return 0;
        //Judge
        int n=height.size();
        //store
        int *left=new int[n];
        int *right=new int[n];
        int maxArea=0;
        //left
        left[0]=1;
        for(int i=1;i<n;i++)
        {
            if(height[i]>height[i-1])
            {
                left[i]=1;
            }
            else
            {
                int j=i-1;
　　　　　　　　　　//always easy miss =
                while(j>=0&&height[j]>=height[i])
                {
                    j-=left[j];
                }
                left[i]=i-j;
                //cout<<"i"<<left[i]<<endl;
            }
        }
        //right
        right[n-1]=1;
        for(int i=n-2;i>=0;i--)
        {
            if(height[i]>height[i+1])
            {
                right[i]=1;
            }
            else
            {
                int j=i+1;
                while(j<n&&height[j]>=height[i])
                {
                    j+=right[j];
                }
                right[i]=j-i;
            }
        }
        //MaxArea
        for(int i=0;i<n;i++)
        {
            //former problem cannot use right+left,1.left is not abs 2.i is not 0
            maxArea=max(maxArea,(right[i]-left[i]+1)*height[i]);
            cout<<"r"<<right[i]<<"l"<<left[i]<<endl;
            cout<<"M"<<maxArea<<endl;
        }
        return maxArea;
    }
};
int main()
{
    vector<int> res={
     1,1};
    Solution s;
    cout<<s.largestRectangleArea(res)<<endl;
    return 0;
}

2.O(n^2)

省代码但是耗时

#include<stdio.h>
#include<iostream>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<algorithm>
#include<stack>
#include<climits>
//
using namespace std;
//similar to the idea of scrolling array
class Solution
{
public:
    int largestRectangleArea(vector<int>& height)
    {
        //NULL
        if(height.empty()) return 0;
        //initialize
        //Judge
        int n=height.size();
        //store
        int minH=INT_MAX;
        int maxArea=0;
        //main part
        //forward
        for(int i=0;i<n;i++)
        {
            if(i+1<n&&height[i]<=height[i+1])
            {
                continue;
            }
            minH=INT_MAX;
            //backward
            for(int j=i;j>=0;j--)
            {
                minH=min(minH,height[j]);
                cout<<"mH     "<<minH<<"      h[j]     "<<height[j]<<endl;
                int area=(i-j+1)*minH;
                maxArea=max(maxArea,area);
                cout<<" minH "<<minH<<" i "<<i<<" j "<<j<<" maxArea "<<maxArea<<endl;
            }
        }
        //return
        return maxArea;
    }
};
int main()
{
    vector<int> res={
     2,1,5,6,2,3};
    Solution s;
    cout<<s.largestRectangleArea(res)<<endl;
    return 0;
}

单调栈：

C++：

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        int res = 0;
        for (int i = 0; i < height.size(); ++i) {
            if (i + 1 < height.size() && height[i] <= height[i + 1]) {
                continue;
            }
            int minH = height[i];
            for (int j = i; j >= 0; --j) {
                minH = min(minH, height[j]);
                int area = minH * (i - j + 1);
                res = max(res, area);
            }
        }
        return res;
    }
};

python：

class Solution(object):
    def largestRectangleArea(self, heights):
        """
        :type heights: List[int]
        :rtype: int
        """
        # store
        stack = list()
        res = 0
        heights.append(0)
        # judge
        N = len(heights)
        for i in range(N):
            if not stack or heights[i] > heights[stack[-1]]:
                stack.append(i)
            else:
                while stack and heights[i] <= heights[stack[-1]]:
                    h = heights[stack[-1]]
                    # pop to find boundary
                    stack.pop()
                    # special minus distance i distance+1 stack[-1] distance-1
                    w = i if not stack else i - stack[-1] - 1
                    res = max(res, h * w)
                # this step easy forget.i isn't former calculation.Need to push stack
                stack.append(i)
        return res
solu=Solution()
# 赋值输出一条龙
print(solu.largestRectangleArea(heights=[2,1,5,6,2,3]))