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.X轴为底，，两个纵轴为变，求容器的容积，短边是瓶颈。
thinking:
(1)短边决定水箱的有效高，底要尽可能的宽。
（2）典型的双指针求解的题型。
（3）贪心的策略，哪条边短，往里收缩寻找下一条边。
code:
class Solution {public:int maxArea(vector<int> &height) {// Start typing your C/C++ solution below// DO NOT write int main() functionint i = 0;int j = height.size() – 1;int ret = 0;while(i < j){int area = (j – i) * min(height[i], height[j]);ret = max(ret, area);if (height[i] <= height[j])i++;elsej–;}return ret;}};
时间复杂度为O（n）
暴力破解法：时间复杂度为O（n*n）
int area(vector<int>::iterator &a,vector<int>::iterator &b){return (b-a)*(*a>*b?*b:*a);}class Solution {public:int maxArea(vector<int> &height) {int max_area=0;for(vector<int>::iterator i=height.begin()+1;i!=height.end();i++){for(vector<int>::iterator j=height.begin();j!=i;j++){max_area=max(max_area,area(j,i));}}return max_area;}};
生活若剥去了理想、梦想、幻想，那生命便只是一堆空架子
