Container with most water:

GivenNnon-negative integersa1 ,a2 , ...,an , where each represents a point at coordinate (I,ai ).NVertical lines is drawn such that the both endpoints of lineIis at (I,ai ) and (I, 0). Find lines, which together with X-axis forms a container, such that the container contains the most water.

Note:you may not slant the container.

Interpretation of the topic:

There are n non-negative numbers, representing the height of the N water barrier plate, now for the maximum capacity of two water-barrier, that is, to find a pair of water-plate, distance *min (two water barrier height) maximum.

Problem-Solving ideas: Method One: The code of the poor lifting method is as follows:

Class Solution {public:    int Maxarea (vector<int>& height) {        int length=height.size ();        if (length<=1) return 0;        int cap=0;        for (int i=0;i<length-1;i++)        {for            (int j=i+1;j<length;j++)            {                if (min (Height[i],height[j]) * ( j-i) >cap)                cap=min (Height[i],height[j]) * (j-i);            }        }        return cap;    }};

The method is straightforward, but there is a timeout. Method Two: The maximum distance of two partitions of the water storage, when the distance is reduced, want to increase the amount of water, the minimum height of the partition should be increased accordingly. Using this idea, we have the following code:

Class Solution {public:    int Maxarea (vector<int>& height) {        int length=height.size ();        if (length<=1) return 0;        int cap=0;        int left=0,right=length-1;        while (Left<right)        {            if (min (Height[left],height[right]) * (right-left) >cap)            Cap=min (height[left ],height[right]) * (right-left);            if (Height[left]



Container with most water: