LeetCode刷题笔录Container With Most Water

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.

只想出了O(n^2)的brute force解法。google了一下发现了这么个解法：

start 和end分别指向数组的头和尾。则当前的container的容积为area=min(height[start],height[end])*(end-start)。假设height[start]<height[end],那么无论end取任何一个值，以start为一条边的容器的area不可能比当前的这个area更大。原因是容器的容积是由较短的那条边决定的。因此这是做start++，进行下一次比较。对于height[end]<height[start]的情况同理end--。

public class Solution {
public int maxArea(int[] height) {
if(height.length < 2)
return 0;
int start = 0;
int end = height.length - 1;
int maxArea = 0;
int area;
while(start < end){
if(height[start] < height[end]){
area = height[start] * (end - start);
start++;
}
else{
area = height[end] * (end - start);
end--;
}
if(maxArea < area)
maxArea = area;
}
return maxArea;
}
}