[LeetCode] Container With Most Water

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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.

[LeetCode] Container With Most Water

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原文地址：http://www.cnblogs.com/jcliBlogger/p/4646659.html