【LeetCode】084. Largest Rectangle in Histogram

题目：

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.

 

For example,
Given heights = [2,1,5,6,2,3],
return 10.

题解：

Solution  1 

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        if(heights.size() < 1)
            return 0;
        return maxArea(heights, 0, heights.size() - 1);
    }
private:
    int combineArea(vector<int> &heights, int l, int m, int r){
        int i = m, j = m;
        int area = 0, h = min(heights[i], heights[j]);
        while(i >= l && j <= r){
            h = min(h, min(heights[i], heights[j]));
            area = max(area, h * (j - i + 1));
            if(i == l)
                ++j;
            else if(j == r)
                --i;
            else {
                if(heights[i - 1] > heights[j + 1])
                    --i;
                else
                    ++j;
            }
        }
        
        return area;
    }
                       
    int maxArea(vector<int> &heights, int l, int r){
        if(l >= r)
            return heights[l];
        int m = l + (r - l) / 2;
        int area = maxArea(heights, l, m - 1);
        area = max(area, maxArea(heights, m + 1, r));
        area = max(area, combineArea(heights, l, m, r));
        return area;
    }
};

 

Solution  2 摘自geeks ，用到了单调栈的思想

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        int res = 0, area_top = 0;
        stack<int> s;
        int n = heights.size();
        int i = 0;
        for(i = 0; i < n;){
            if(s.empty() || heights[s.top()] < heights[i]){
                s.push(i++);
            } else {
                int cur = s.top();s.pop();
                area_top = heights[cur] * (s.empty() ? i : i - s.top() - 1); 
                res = max(res, area_top);
            }
        }
        while(!s.empty()){
            int cur = s.top();s.pop();
            area_top = heights[cur] * (s.empty() ? i : i - s.top() - 1);
            res = max(res, area_top);
        }
        return res;
    }
};

 

Solution  3 优化版

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        int res = 0;
        stack<int> s;
        int n = heights.size();
        heights.push_back(0);
        for(int i = 0; i <= n;){
            if(s.empty() || heights[s.top()] < heights[i]){
                s.push(i++);
            } else {
                int cur = s.top();s.pop();
                int area_top = heights[cur] * (s.empty() ? i : i - s.top() - 1); 
                res = max(res, area_top);
            }
        }
        return res;
    }
};

 

Solutin 4 实际上有些类似，不过把栈的空间复杂度转化为求面积时的时间复杂度，实际上花费时间要多，不推荐此做法。

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

 

Solution 4 摘自geeks  基于线段树