标签：leetcode   算法   

问题描述：

Given n non-negative integers a₁, a₂, ..., a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) 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.

基本思路：

此题最简单的想法是遍历每种可能。时间复杂度是O(n^2);

更好的一个思路是每次改变容器左右壁的短板，希求能通过提升短板的高度来找到更大的容量。可以结合代码理解思路。

代码：

int maxArea(vector<int> &height) {  //C++
        int begin = 0 ;
        int end = height.size()-1;
        int max = 0, tmpArea;
        
        while(begin < end){
            tmpArea = (end-begin)*((height[begin]>height[end])?height[end]:height[begin]);
            if(tmpArea > max){
                max = tmpArea;
            }
            if(height[begin] > height[end])
                end--;
            else begin++;
        }
        return max;
    }

[leetcode]Container With Most Water

标签：leetcode   算法   

原文地址：http://blog.csdn.net/chenlei0630/article/details/41089517