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.
思路：这道题就是求最大矩形面积。当长度减少时，高度必须增加才能增大面积。使用两个指针——首指针和尾指针,取min(ai,aj)*(j-i).如果头指针指向的结点小于尾指针指向的结点，这头指针向前移动，否则尾指针向前移动，再继续求解矩形面积。感觉就像贪心算法，如果距离减小，必须通过增加高度来弥补。
class Solution { public: int maxArea(vector<int> &height) { int nSize=height.size(); if(nSize==0) return 0; int Area=0; int mostWater=0; vector<int>::iterator iter1=height.begin(); vector<int>::iterator iter2=height.end()-1; while(iter1<iter2) { Area=min(*iter1,*iter2)*(iter2-iter1); if(mostWater<Area) mostWater=Area; if(*iter1<*iter2) iter1++; else iter2--; } return mostWater; } };
Container With Most Water,布布扣,bubuko.com
原文：http://www.cnblogs.com/awy-blog/p/3619989.html