[LeetCode] Container With Most Water

最新推荐文章于 2020-08-23 15:53:30 发布

转载最新推荐文章于 2020-08-23 15:53:30 发布 · 54 阅读

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原文链接：http://www.cnblogs.com/jcliBlogger/p/4646659.html

本文探讨了如何使用两个指针解决求解容器中最大水量的问题，并提供了两种不同的C++实现方式。通过从两端向中间逼近的方式，有效地将复杂度控制在O(n)内。

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Well, an interesting problem. If you draw some examples on a white board and try to figure out the regularities, you may have noticed that the key to solving this problem is to use two pointers, one starting from the left end and the other starting from the right end. We compute an area for one configuration of the two pointers. Then we need to move them. If the left pointer has lower height, then we know it must be moved to one point with larger height in the right to make it possible for the area to increase. For the right pointer, the case is similar.

The following code is not well optimized, but being pretty short :-)

 1 class Solution {
 2 public:
 3     int maxArea(vector<int> &height) {
 4         int l = 0, r = height.size() - 1, area = 0;
 5         while(l < r) {
 6             area = max(area, min(height[r], height[l]) * (r - l));  
 7             height[l] <= height[r] ? l++ : r--;
 8         }
 9         return area;
10     }
11 };

You may avoid unnecessary checks by moving the left and right pointers to another place with larger height. However, the following code runs slower than the above one...

 1 class Solution {
 2 public:
 3     int maxArea(vector<int> &height) {
 4         int l = 0, r = height.size() - 1, area = 0;
 5         while(l < r) {
 6             int h = min(height[l], height[r]), w = r - l;
 7             area = max(area, h * w);  
 8             while (height[l] <= h && l < r) l++;
 9             while (height[r] <= h && r > l) r--;
10         }
11         return area;
12     }
13 };

Since the problem is not that hard, you may be required to think more, such at its time complexity. Could you prove that it runs in O(n) time? Well, this link has a very clever proof.

转载于:https://www.cnblogs.com/jcliBlogger/p/4646659.html