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 and n is at least 2.

思路：双指针，贪心策略。先确定扫描长度是最长的，也就是从两端扫描起，找到两端较短的木板后，往中间靠拢，使短板变长，再计算更新最大面积，当双指针在中间汇集时结束。
木桶效应：装水的面积由短板决定。

class Solution {public:int maxArea(vector<int>& height) {    int water = 0;    int i = 0, j = height.size() - 1;    while (i < j) {        int h = min(height[i], height[j]);        water = max(water, (j - i) * h);        while (height[i] <= h && i < j) i++;        while (height[j] <= h && i < j) j--;    }    return water;}};