Medium 11题 Container With Most Water

Question:

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.

Solution:

This method is too stupid....... and it is TLE...

public class Solution {
public int maxArea(int[] height) {
int n=height.length;
int max=0;
for(int i=0;i<=n-1;i++)
{
for(int j=i+1;j<=n-1;j++)
{
// if(max>((j-i)*(height[i]>height[j]?height[j]:height[i]))
// max=(j-i)*(height[i]>height[j]?height[j]:height[i]);
max=max>((j-i)*(height[i]>height[j]?height[j]:height[i]))?max:(j-i)*(height[i]>height[j]?height[j]:height[i]);
}
}
return max;
}
}

So...maybe we must analysis the number...

设置两个指针i, j, 一头一尾, 相向而行. 假设i指向的挡板较低, j指向的挡板较高(height[i] < height[j]).
下一步移动哪个指针?
-- 若移动j, 无论height[j-1]是何种高度, 形成的面积都小于之前的面积.
-- 若移动i, 若height[i+1] <= height[i], 面积一定缩小; 但若height[i+1] > height[i], 面积则有可能增大.
综上, 应该移动指向较低挡板的那个指针.

public class Solution {
public int maxArea(int[] height) {
int n=height.length;
int i=0;
int j=n-1;
int max=(j-0)*(height[0]>height[j]?height[j]:height[0]);
while(i<j)
{
int ans=(j-i)*(height[i]>height[j]?height[j]:height[i]);
max=max>ans?max:ans;
if(height[i]<height[j])
i++;
else
j--;
}
return max;
}
}

还是有点笨？////