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;
    }
}