11. Container With Most Water
Medium
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 and n is at least 2.
The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
Example:
Input: [1,8,6,2,5,4,8,3,7]
Output: 49
题目链接：https://leetcode.com/problems/container-with-most-water/
解题思路：
原本按照算法树的思路，想从动态规划入手，从后往前找满足所需宽度距和高度要求的木板，企图从暴力的O(n^2)尽量向O(n)靠，但没能找到限制向前查找的停止条件。
后来看了solution才恍然大悟用双指针竟然如此简单。
方法：双指针向中间夹逼 O(n)
这里有一条规律：组成最大容积的两块板l和r，l的左边没有高度大等于l的板，r右边没有高度大等于r的板。这个规律很容易用数学式子证明。因此可以用双指针夹逼，从宽最大的情况，向中间找时，以高度换宽度。
具体步骤：两个指针从两头向中间靠拢，同时更新最大容积；在收缩区间的时候优先从中较小的边开始收缩。
代码：
class Solution {
public:
int maxArea(vector<int>& height) {
int l = 0, r = height.size() - 1, maxval = 0;
while (l < r){
maxval = max(maxval, min(height[l], height[r]) * (r - l));
(height[l] <= height[r])? l++: r--;
}
return maxval;
}
};
一个小总结：
对找区间的题，可以考虑用双指针。双指针有从同一起点出发，有两边向中间夹逼。
本篇参考：https://blog.csdn.net/a83610312/article/details/8548519
http://www.cnblogs.com/grandyang/p/4455109.html