[Leetcode] Container With Most Water (Java)

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.

题意就是在坐标轴上找两条线x=l,x=r,使(r-l)*min(height[l],height[r])最大

从两边往中间找，不停找比当前线更长的线，计算面积大小，若比当前最大面积大，则更新，然后继续找，直到两边碰头

public class ContainerWithMostWater {
public int maxArea(int[] height) {

int lindex = 0;
int rindex = height.length-1;
int max = (rindex-lindex)*Math.min(height[lindex],height[rindex]);
while(rindex>lindex) {
if(height[lindex]<height[rindex]) {
int k = lindex+1;
while(height[k]<height[lindex]&&k<rindex)
k++;
lindex=k;
}else {
int k = rindex-1;
while(height[k]<height[rindex]&&k>lindex)
k--;
rindex=k;
}
int temp = (rindex-lindex)*Math.min(height[rindex],height[lindex]);
if(max<temp){
max=temp;
}
}
return max;
}
public static void main(String[] args) {
int[] height = {3,2,1,3};
System.out.println(new ContainerWithMostWater().maxArea(height));
}
}