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;
     }
 };