[Leetcode] 11. Container With Most Water

Problem

Given n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

思路：
这里相当于求最大矩形面积。这里有一个tradeoff，一个是横轴的长度（距离），一个是纵轴的高度。
1. 暴力地做两层循环，遍历其中两两元素，求最大的矩形面积，时间复杂度为O(n^2)。这种暴力的方法当然是TLE的……
2. 动态规划。先对高度进行从小到大排序，然后从高度最小的元素开始访问。计算离其最远的X轴距离，得到矩形面积。此后便不再考虑这个元素（短板）。这样的方法时间复杂度是O(nlogn)(排序)+O(n)(访问各个元素)。
3. 【ref】双指针法。分别分配两个指针在最左和最右，计算矩形面积。如果height[left]

def calArea(i,j,height):    height = min(height[i],height[j])    area = (j-i)*height    return areaclass Solution(object):    def maxArea(self, height):        """        :type height: List[int]        :rtype: int        """        hlen = len(height)        i = 0        maxarea = 0;        while i < hlen:            j = i + 1            while j < hlen:                tmparea = calArea(i,j,height)                maxarea = maxarea if maxarea > tmparea else tmparea                j = j + 1            i = i + 1        return maxarea

Solution 2: [Accepted] 4%

class Solution(object):    def maxArea(self, height):        """        #:type height: List[int]        #:rtype: int        """        hlen = len(height)        l = zip(xrange(0,hlen),height)        newdic =  sorted(l,key = lambda d:d[1],reverse=False)        i = 0        maxarea = 0        xmax = hlen - 1        xmin = 0        flag =  [ True for n in xrange(hlen)]        while i < hlen-1:            while flag[xmax] == False:                xmax = xmax - 1            while flag[xmin] == False:                xmin = xmin + 1            x1 = xmax-newdic[i][0]            x2 = newdic[i][0]-xmin            delta = x1 if x1 > x2 else x2            tmparea = delta * newdic[i][1]            maxarea = maxarea if maxarea > tmparea else  tmparea            flag[newdic[i][0]] = False            i = i + 1        return maxarea

ref:
python内置方法的时间复杂度