Leetcode container with the most water and trapping Rain water

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.

Analysis reference here, the author explains it in detail.

Class Solution {public:    int. maxarea (vector<int> &height) {    int left = 0,right = Height.size () -1,area = 0 ;    while (left < right)    {    area = max (Area,min (Height[left],height[right]) * (right-left));//Current Location    if ( Height[left] < Height[right])    {    int k = left;    while (K < right && Height[k] <= height[left]) ++k;//find the next position left    = k;    }    else    {    int k = right;    while (K > left && height[k] <= height[right])--k;//Find Next position right    = k;    }    }    return area;    }};

Trapping Rain Water

given n  non-negative integers representing an elevation map where The width of each bar was 1, compute how much water it was able to trap after raining.

For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1] , return 6 .

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of the Rain Water (blue section) is being trapped. Thanks Marcos for contributing this image!

See here, the author explained in detail

Class Solution {public:    int trap (int a[], int n)     {    vector<int> leftmostheight (n,0), Rightmostheight ( n,0);    int mostheiht = 0,i,area = 0;    for (i = 0;i < N;++i)    {    Leftmostheight[i] = mostheiht;//to the left of the maximum height    mostheiht = max (Mostheiht,a[i]);    }    mostheiht = 0;    for (i = n-1;i >= 0;--i)    {    Rightmostheight[i] = mostheiht;//to the right of the maximum height    mostheiht = max (Mostheiht,a[i]); c14/>}    for (i = 0;i < N;++i)    {    int a = min (Leftmostheight[i],rightmostheight[i])-a[i];//The current height of the    if (a > 0) area + = A;    }    return area;    }};

There are also two similar topics, all of which are related to histograms, see here

Leetcode container with the most water and trapping Rain water