小白笔记-------------------------------leetcode(11. Container With Most Water )

Given n non-negative integers a₁, a₂, ...,a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of linei is at (i, a_i) 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.

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普通思路就是两个for循环全部遍历，但是会超时，这里需要考虑的就是剪枝的问题，那么如何减枝？

很容易想到的方法就是类似与左右子树的减枝，就是用两个游标，左右游标来控制，

那么定义了左右游标之后，如何保证最大的值仍能被遍历到呢？

首先，将值与最大值比较，保存

接着，因为减大值的话没有减枝的作用，所以选择小的一端往里进一格，至此问题就解决了。

class Solution {    public int maxArea(int[] height) {        int maxArea = 0;//初始化一个最大值        int Area = 0;        int minHeight = 0;        int n = height.length;        int left = 0;        int right = n-1;        while(left < right){            minHeight = height[left] <= height[right]?height[left]:height[right];            Area = (right-left) * minHeight;            if(Area > maxArea){                maxArea = Area;            }            if(height[left] < height[right]){                left++;            }else{                right--;            }                    }                                return maxArea;    }}