Leetcode Container with most water

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 and 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.

For this problem, the first is to use the method of violence, the complexity of O (N2),

1  PackageContainer.With.Most.Water;2 3  Public classContainerwithmostwater {4 5     /**6      * @paramargs7      */8      Public Static voidMain (string[] args) {9         //TODO auto-generated Method StubTen        intheight[]={1,2,3,4}; One        intMaxarea=maxarea (height); A System.out.println (Maxarea); -     } -      the      Public Static intMaxarea (int[] height) { -        intMaxarea=0; -         for(inti=0;i){ -             for(intj=i+1;j){ +                intminheight=math.min (Height[i], height[j]); -                intkuan=j-i; +                inttemparea=kuan*MinHeight; A                if(temparea>Maxarea) atMaxarea=Temparea; -            } -        } -        returnMaxarea; -     } -  in}

This situation directly timed out, then according to the nature of the problem can take a linear complexity method

1. First, suppose we find the longitudinal line that can take the maximum volume as I, J (assuming I<j), then the maximum volume C = min (ai, AJ) * (j-i);

2. Let's look at a property like this:

①: No line at the right end of J will be higher than it! Hypothesis exists K | (j<k && ak > AJ), then by ak> aj, so min (Ai,aj, AK) =min (AI,AJ), so the volume of the container consisting of I, k c ' = min (ai,aj) * (k-i) > C, and C is the most value contradiction, so the proof of J will not have a higher line than it;

②: Similarly, there will be no higher line on the left side of I;

What does that mean? If we currently get the candidate: set to X, y two lines (x< y), then be able to get a larger volume than its new two edges necessarily within the [x, Y] interval and ax ' > =ax, ay ' >= ay;

3. So we move from the two to the middle, while updating the candidate values, in the contraction interval priority from the X, y of the smaller edge of the contraction;

Start at both ends and gradually shrink

1  PackageContainer.With.Most.Water;2 3  Public classContainerWithMostWater1 {4 5     /**6      * @paramargs7      */8      Public Static voidMain (string[] args) {9         //TODO auto-generated Method StubTen            intheight[]={1,2,3,4}; One            intMaxarea=maxarea (height); A System.out.println (Maxarea);  -     } -       Public Static intMaxarea (int[] height) { the          intMaxarea=0; -          intI=0; -          intJ=height.length-1; -           while(i!=j) { +              intTemparea= (j-i) *math.min (Height[i], height[j]); -              if(temparea>Maxarea) +              { AMaxarea=Temparea; at              } -              if(height[i]<Height[j]) { -i++; -}Else{ -j--; -              } in          } -          returnMaxarea; to      } +}

Leetcode Container with most water