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原题地址：https://oj.leetcode.com/problems/largest-rectangle-in-histogram/
题意：
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 height = [2,1,5,6,2,3],
return 10.
解题思路：又是一道很巧妙的算法题。
Actually, we can decrease the complexity by using stack to keep track of the height and start indexes. Compare the current height with previous one.
Case 1: current > previous (top of height stack)
Push current height and index as candidate rectangle start position.
Case 2: current = previous
Ignore.
Case 3: current < previous
Need keep popping out previous heights, and compute the candidate rectangle with height and width (current index - previous index). Push the height and index to stacks.
(Note: it is better use another different example to walk through the steps, and you will understand it better).
[最优解法] (下面提到的图2是制题目里面的第2个图)
class Solution: # @param height, a list of integer # @return an integer def largestRectangleArea(self, height): #如图2所示，从左到右处理直方，当i = 4 时，小于当前栈顶（即直方3），对于直方3，无论后面还是前面的直方，都不可能得到比目前栈顶元素更高的高度了，处理掉直方3（计算从直方3到直方4 之间的矩形的面积，然后从栈里弹出）；对于直方2 也是如此；直到碰到比直方4 更矮的直方1。这就意味着，可以维护一个递增的栈，每次比较栈顶与当前元素。如果当前元素小于栈顶元素，则入栈，否则合并现有栈，直至栈顶元素小于当前元素。 stack, area, i = [], 0, 0 height.append(0) while i < len(height): if stack == [] or height[i] > height[stack[-1]]: stack.append(i) i += 1 else: curr = stack.pop() width = i if stack == [] else i - (stack[-1] + 1) area = max(area, height[curr] * width) return area
这个是别人的解法，但是过于累赘。
class Solution: # @param height, a list of integer # @return an integer # @good solution! def largestRectangleArea(self, height): stack=[]; i=0; area=0 while i<len(height): if stack==[] or height[i]>height[stack[len(stack)-1]]: stack.append(i) else: curr=stack.pop() width=i if stack==[] else i-stack[len(stack)-1]-1 area=max(area,width*height[curr]) i-=1 i+=1 while stack!=[]: curr=stack.pop() width=i if stack==[] else len(height)-stack[len(stack)-1]-1 area=max(area,width*height[curr]) return area
常规解法，所有的面积都算一遍，时间复杂度O(N^2)。不过会TLE。
class Solution: # @param height, a list of integer # @return an integer # @good solution! def largestRectangleArea(self, height): maxarea=0 for i in range(len(height)): min = height[i] for j in range(i, len(height)): if height[j] < min: min = height[j] if min*(j-i+1) > maxarea: maxarea = min*(j-i+1) return maxarea
[leetcode]Maximal Rectangle @ Python [图解] [很难]
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原文地址：http://www.cnblogs.com/asrman/p/4001381.html
