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11. Container With Most Water

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작성일 2022.04.05 12:56

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[Leetcode 시즌 3] 2022년 4월 5일 문제입니다.

[Medium] 11. Container With Most Water

You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).

Find two lines that together with the x-axis form a container, such that the container contains the most water.

Return the maximum amount of water a container can store.

Notice that you may not slant the container.

Example 1:

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.

Example 2:

Input: height = [1,1]
Output: 1

Constraints:

n == height.length
2 <= n <= 105
0 <= height[i] <= 104

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링크

https://leetcode.com/problems/container-with-most-water/ 3888 회 연결

익명
작성일 2022.04.05 22:19

리트코드 많이 풀어보지 않아서 하루에 한개씩 풀어보려고 ... .
def maxArea(self, height: List[int]) -> int:
max_container, left, right = 0, 0, len(height) -1
while left < right:
cur_container = (right - left) * min (height[left], height[right])
max_container = max (max_container, cur_container)
if height[left] < height[right] :
left += 1
else:
right -= 1
return max_container

익명
작성일 2022.04.06 15:42

응원합니다. 하루에 한개씩 한달이면 30개~ 1년이면 360개~ 화이팅입니다.

익명
작성일 2022.04.06 08:58

from collections import deque
class Solution:
    def maxArea(self, height: List[int]) -> int:
        m, q = 0, deque(height)
        while q:
            m = max(m, min(q[0], q[-1]) * (len(q) - 1))
            if q[0] > q[-1]:
                q.pop()
            else:
                q.popleft()
        return m

익명
작성일 2022.04.06 09:37

class Solution {
public:
    int maxArea(vector<int>& height) {
        int maxAreaValue=0;
        int first=0;
        int last=height.size()-1;
        
        while(last>first){
                if( (last-first)*(height[first] > height[last] ? height[last] : height[first]) > maxAreaValue){
                    maxAreaValue=(last-first)*(height[first] > height[last] ? height[last] : height[first]);
                };
                if(height[first]>height[last]){
                    last--;
                } else {
                    first++;
                };
        };
        return maxAreaValue;
    }
};

익명
작성일 2022.04.06 15:43

문제 풀이 감사합니다.

익명
작성일 2022.04.06 21:10

class Solution:
    def maxArea(self, height: List[int]) -> int:
        left = 0
        right = len(height)-1
        maxArea = 0
        
        while left < right:
            area = (min(height[left], height[right]))*(right-left)
            if area > maxArea:
                maxArea = area
            else:
                if height[left] < height[right]:
                    left += 1
                else:
                    right -= 1
            
        return maxArea

익명
작성일 2022.04.14 21:24

class Solution {
    func maxArea(_ height: [Int]) -> Int {        
        var left = 0
        var right = height.count-1
        var area = 0
        while right > left {
            let hei = min(height[left], height[right])
            let wid = right - left
            let newArea = wid * hei 

            if newArea > area {
                area = newArea
            }
            
            if height[left] > height[right] {
                right -= 1
            } else {
                left += 1
            }
        }
        
        return area
    }
}

익명
작성일 2022.04.17 13:24

class Solution:
    def maxArea(self, height: List[int]) -> int:
        p1, p2, ans = 0, len(height) - 1, 0
        while p1 < p2:
            temp = 0
            if height[p1] < height[p2]:
                temp = (p2 - p1) * height[p1]
                p1 += 1
            else:
                temp = (p2 - p1) * height[p2]
                p2 -= 1
            ans = max(ans, temp)
        return ans