LeetCode-42-Trapping Rain Water 贪心或单调栈

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这个题一开始想麻烦了，我用单调栈实现了一下，A了，但是只打败了0.4%的人，后来看了一眼别人的博客，发现没那么麻烦，找到最高点，从左右贪心就行了。。。

下面是两份不同做法的代码，时间复杂度都是O（n）

class Solution(object):    def trap(self, height):        """        :type height: List[int]        :rtype: int        """        Len=len(height)        if Len<=1:return 0        ans=0        MAX=0        p=0        for i in range(0,Len):            if height[i]>MAX:                MAX=height[i]                p=i        cur=height[0]        for i in range(1,p):            if cur<height[i]:                cur=height[i]            else:                ans+=cur-height[i]                cur=height[Len-1]        for i in range(Len-2,p,-1):            if cur<height[i]:                cur=height[i]            else:                ans+=cur-height[i]        return ans

class Solution(object):    def trap(self, height):        """        :type height: List[int]        :rtype: int        """        Len=len(height)        if Len<=1:return 0        stack=[]        stack.append((height[0],0))        ans=0        #print "len",Len        for i in range(1,Len):            print i            stackSize=len(stack)            if height[i]<stack[stackSize-1][0]:                #print"fuck1"                stack.append((height[i],i))                print stack            elif height[i]==stack[stackSize-1][0]:                #print "fuck2"                stack.pop()                stack.append((height[i],i))            else:                #print"fuck"                hereH=height[i]                V=0                lowest=stack.pop()                V+=lowest[0]                stackSize-=1                while(stackSize>0 and stack[stackSize-1]<=lowest[0]):                    lowest=stack.pop()                    V+=lowest[0]                    stackSize-=1                higher=(-1,-1)                V-=lowest[0]                while stackSize>0 and stack[stackSize-1]>=lowest[0]:                    higher=stack.pop()                    V+=lowest[0]*(lowest[1]-higher[1])                    #print "V",V                    stackSize-=1                    if higher[0]>=hereH:                        stack.append(higher)                        break                    lowest=higher                if higher[0]!=-1:                    ans+=min(higher[0],hereH)*(i-higher[1]-1)-V                    #print "ans",ans,V                    if higher[0]==hereH:                        stack.pop()                stack.append((hereH,i))        return ans

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