Leetcode 盛水最多的容器

算法思路：

Explanation:

• Two-pointer technique:

  • Start with two pointers, one at the beginning (left) and one at the end (right) of the height array.
  • Calculate the area formed between the two vertical lines. The area is calculated as: $$\text{Area}=(\text{right}-\text{left})\times \min (\text{height[left]},\text{height[right]})$$
  • Move the pointer that points to the shorter line inward (either left or right), as moving the shorter line has the potential to increase the maximum area.
  • Repeat this process until the two pointers meet.

• Time Complexity:

  • The solution runs in $O(n)$, where $n$ is the number of elements in the height array, because each element is processed at most once.

代码：

class Solution {
public:
    int maxArea(vector<int>& height) {
        int left = 0;
        int right = height.size() - 1;
        int maxWater = 0;
        while(left < right) {
            int currentWater = (right - left) * min(height[left], height[right]);
            if(height[left] >= height[right]) {
                right--;
            }else {
                left++;
            }
            maxWater = max(maxWater, currentWater);
        }
        return maxWater;
    }
};