[Leetcode]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.

题意：给定一系列和x轴垂直相交的线，线的起点坐标和终点坐标分别为(i, ai)和
(i,
0)，找出其中的两条线满足Min(ai, aj)
* |j - i|最大(j和i代表垂直线的x轴值，j和i作为角标，谁大谁小都可以，这里假定j > i)。

解法一：时间复杂度为O(N^2)，双重循环解法本文不做讨论；

解法二：时间复杂度为O(N)

public class Solution {
public int maxArea(int[] height) {
int left = 0;
int right = height.length - 1;

int maxArea = 0;
int currentArea = 0;
while (left < right) {
currentArea = (right - left) * (height[left] < height[right] ? height[left] : height[right]);

if (currentArea > maxArea) {
maxArea = currentArea;
}

if (height[left] <= height[right]) {
++left;
} else {
--right;
}
}
return maxArea;
}
}

对于两条线：(i, Ai)，(j, Aj)，i < j，它们所形成的容器的面积为：
area(i, j) = (j - i) * min{Ai, Aj}
当Ai < Aj时，对于以(i, Ai)为一条垂直边的所有容器，若在[i, j]直接有另外一条线[k, Ak]，[i, Ai]和[k, Ak]形成的容器面积为
area(i, k) = (k - i) * min{Ai, Ak}，
显然 k -i < j -i，所以：
(k
- i) * min{Ai, Ak} < (j - i) * min{Ai, Ak}，
而(k
- i) * min{Ai, Ak} <= (k - i) * Ai < (j - i) * Ai < (j- i) * min{Ai, Aj},

即(k - i) * min{Ai, Ak} < (j-
i) * min{Ai, Aj}，

结论：在[i, j]，且Ai < Aj，在以[i, Ai]为一条边的所有水容器中，[i, Ai]和[j, Ak]所形成的水容器面积最大。