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=0,b=n-1,contain=min(height[0],height[n-1])*(n-1);
        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;
    }
};