小白筆記-------------------------------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 line i 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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普通思路就是兩個for循環全部遍歷，但是會超時，這裏需要考慮的就是剪枝的問題，那麼如何減枝？

很容易想到的方法就是類似與左右子樹的減枝，就是用兩個遊標，左右遊標來控制，

那麼定義了左右遊標之後，如何保證最大的值仍能被遍歷到呢？

首先，將值與最大值比較，保存

接着，因爲減大值的話沒有減枝的作用，所以選擇小的一端往裏進一格，至此問題就解決了。

class Solution {
    public int maxArea(int[] height) {
        int maxArea = 0;//初始化一個最大值
        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;
    }
}