Skip to content

LeetCode: 11-Container With Most Water 解題紀錄

Last Updated on 2022-04-05 by Clay

題目

Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of the line i is at (i, ai) and (i, 0). Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.

Notice that you may not slant the container.

Example:

Input: height = [1,8,6,2,5,4,8,3,7]
Output: 49
Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

這個題目也是相當地經典，就是題目給定一陣列，陣列內為由左到右每個 …… ㄜ，這裡就叫『木板』好了，每個木板的高度。

而不同的木板之間，距離單位都是 1。我們要做的就是，找出能讓容器積蓄最多的水兩片木板，最後，回傳可以積蓄水的最大值。

解題思路

一開始我使用了超級 for 迴圈來遍歷 XDDD

結果就很順利地被 TLE（Time Limit Exceeded）了。

後來想想，這個題目很適合使用像是貪婪演算法（？）這樣的思路去尋找答案。設定左右兩個木板，計算出每一次儲蓄水的大小，然後將比較矮的木板再往下一個位置挪動 —— 也就是盡量保留比較高的木板。

最終，應能求得最大的儲蓄水位。


C++ 程式碼

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


Python 程式碼

class Solution:
    def maxArea(self, height) -> int:
        max_area = 0
        left = 0
        right = len(height)-1

        while left < right:
            temp = min(height[left], height[right]) * (right-left)
            max_area = max(temp, max_area)

            if height[left] < height[right]:
                left += 1
            elif height[left] > height[right]:
                right -= 1
            else:
                left += 1
                right -= 1

        return max_area


References

Tags:

Leave a Reply