[LeetCode] 011. Container With Most Water (Medium) (C ++/Java/Python), leetcodecontainer

[LeetCode] 011. Container With Most Water (Medium) (C ++/Java/Python), leetcodecontainer

Index: [LeetCode] Leetcode index (C ++/Java/Python/SQL)
Github: https://github.com/illuz/leetcode

011. Container_With_Most_Water (Medium) Link:

Title: https://oj.leetcode.com/problems/container-with-most-water/
Code (github): https://github.com/illuz/leetcode

Question:

For some baffle plates, select two baffle plates to find the maximum storage capacity.

Analysis:

Let's look at the detailed algorithm of this great god...

The second algorithm is used here.

Code: C ++:

class Solution {public:    int maxArea(vector<int> &height) {int lpoint = 0, rpoint = height.size() - 1;int area = 0;while (lpoint < rpoint) {area = max(area, min(height[lpoint], height[rpoint]) *(rpoint - lpoint));if (height[lpoint] > height[rpoint])rpoint--;elselpoint++;}return area;    }};

Java:

public class Solution {    public int maxArea(int[] height) {        int lpoint = 0, rpoint = height.length - 1;        int area = 0;        while (lpoint < rpoint) {            area = Math.max(area, Math.min(height[lpoint], height[rpoint]) *                    (rpoint - lpoint));            if (height[lpoint] > height[rpoint])                rpoint--;            else                lpoint++;        }        return area;    }}

Python:

class Solution:    # @return an integer    def maxArea(self, height):        lp, rp = 0, len(height) - 1        area = 0        while lp < rp:            area = max(area, min(height[lp], height[rp]) * (rp - lp))            if height[lp] > height[rp]:                rp -= 1            else:                lp += 1        return area