[leetcode] Largest Rectangle in Histogram @ Python [图解] [很难]

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i=3），对于直方i=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i=3（计算从直方i=3到直方i=4 之间的矩形的面积，然后从栈里弹出）；对于直方i=2 也是如此；直到碰到比直方i=4 更矮的直方i=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。如果当前元素大于栈顶元素，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 - 1) - stack[-1]  #Note; i – 1 resulted from i += 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