84. 柱状图中最大的矩形

分治法，但PYTHON还是超时，因为有个巨长的有序测试数据，此时分治变O(N2)过不去。。

class Solution:
    def largestRectangleArea(self, heights) -> int:
        return recur(heights,0,len(heights))
def recur(heights,le,ri):
    if le>=ri:
        return 0
    mi=le
    for i in range(le,ri):
        if heights[i]<heights[mi]:
            mi=i
    print(le,mi,ri)
    cur_area=(ri-le)*heights[mi]
    if mi>le and mi+1<ri:
        return max(recur(heights,le,mi),recur(heights,mi+1,ri),cur_area)
    if mi>le:
        return max(recur(heights,le,mi),cur_area)
    if mi+1<ri:
        return max(recur(heights,mi+1,ri),cur_area)
    return cur_area

递增栈，主要看的这篇博客

https://blog.csdn.net/qq_17550379/article/details/85093224

class Solution:
    def largestRectangleArea(self, heights) -> int:
        stack=list()
        i,res=0,0
        while i<len(heights):
            if not stack or heights[stack[-1]]<=heights[i]:
                stack.append(i)
                i+=1
            else:
                k=stack.pop()
                res=max(heights[k]*((i-stack[-1]-1) if stack else i),res)
        while stack:
            k=stack.pop()
            res=max(heights[k]*((i-stack[-1]-1) if stack else i),res)
        return res