Leetcode (11) The puzzle: Container with the most water

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

Topic:

Given n non-negative integers A1, A2, ..., an, where each represents a point at Coordi Nate (i, ai). N Vertical Lines is drawn such that the the both endpoints of line I am 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.

Ideas:

Consider reducing the range from the border to the inside. It is important to note that, assuming that the height of the boundary now is a and B, then there are other points in the interior that can be larger than A and B frames, then this point must be larger than the smallest of a and B. So it keeps shrinking from the smaller points of the boundary height until the interior is empty.

Simply iterate through the array once, so the complexity is O (n).

AC Code:

1 classSolution {2  Public:3     intMaxarea (vector<int>&height) {4         intI,j,n=height.size (), a=0, b=n-1, Contain=min (height[0],height[n-1]) * (n1);5          while(a<b) {6             if(height[a]<Height[b]) {7                  for(i=a;i<b;i++){8                     if(height[i]>Height[a])9                          Break;Ten                 } One                 if(min (Height[i],height[b]) * (b-i) >contain) { AContain=min (Height[i],height[b]) * (b-i); -                 } -A=i; the             } -             Else{ -                  for(j=b;j>a;j--){ -                     if(height[j]>Height[b]) +                          Break; -                 } +                 if(min (Height[j],height[a]) * (J-A) >contain) { AContain=min (Height[j],height[a]) * (J-a); at                     } -b=J; -             } -         } -         returncontain; -     } in};

Leetcode (11) The puzzle: Container with the most water