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