[leetcode]Largest Rectangle in Histogram @ Python

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).

代码：

class Solution:
    # @param height, a list of integer
    # @return an integer
    # @good solution!
    def largestRectangleArea(self, height):
        maxArea = 0
        stackHeight = []
        stackIndex = []
        for i in range(len(height)):
            if stackHeight == [] or height[i] > stackHeight[len(stackHeight)-1]:
                stackHeight.append(height[i]); stackIndex.append(i)
            elif height[i] < stackHeight[len(stackHeight)-1]:
                lastIndex = 0
                while stackHeight and height[i] < stackHeight[len(stackHeight)-1]:
                    lastIndex = stackIndex.pop()
                    tempArea = stackHeight.pop() * (i-lastIndex)
                    if maxArea < tempArea: maxArea = tempArea
                stackHeight.append(height[i]); stackIndex.append(lastIndex)
        while stackHeight:
            tempArea = stackHeight.pop() * (len(height) - stackIndex.pop())
            if tempArea > maxArea:
                maxArea = tempArea
        return maxArea

代码：

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。

代码：

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