LeetCode(11)题解: Container With Most Water

https://leetcode.com/problems/container-with-most-water/

题目：

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：

考虑从边界向里缩减范围。最重要的是注意到，假设现在边界的height是a和b，那么想在内部有别的点能比a、b框成的体积大，那么这个点一定得比a、b中的最小者大。所以不断从边界height较小的点向内缩减，直到内部为空。

只需遍历数组一次，所以复杂度为O(n)。

AC代码：

 class Solution {

 public:

     int maxArea(vector<int>& height) {

         int i,j,n=height.size(),a=,b=n-,contain=min(height[],height[n-])*(n-);

         while(a<b){

             if(height[a]<height[b]){

                 for(i=a;i<b;i++){

                     if(height[i]>height[a])

                         break;

                 }

                 if(min(height[i],height[b])*(b-i)>contain){

                     contain=min(height[i],height[b])*(b-i);

                 }

                 a=i;

             }

             else{

                 for(j=b;j>a;j--){

                     if(height[j]>height[b])

                         break;

                 }

                 if(min(height[j],height[a])*(j-a)>contain){

                     contain=min(height[j],height[a])*(j-a);

                     }

                 b=j;

             }

         }

         return contain;

     }

 };

巴特西

LeetCode(11)题解: Container With Most Water

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