Datawhale LeetCode腾讯精选50——Task03

LeetCode11：盛最多水的容器

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.

用时最快的一个解决方案，出自Python版[leetcode]11. 盛最多水的容器（难度中等）。也有另外两个，思路一样，代码实现略微不同，用时稍慢。Leetcode 11.盛最多水的容器 By Python；leetcode 11. 盛最多水的容器 - 两种解法 - python。

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

LeetCode11盛最多水的容器官方解决方案

LeetCode14：最长公共前缀

Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string "".

代码出自Leetcode——14.最长公共前缀（python）。这篇文章给出了三种不同的解法，其中下面这个是用时和占存最少的，思路也很清奇。

class Solution:
    def longestCommonPrefix(self, strs):
        if not strs: return ""
        s1 = min(strs)
        s2 = max(strs)
        for i, x in enumerate(s1):
            if x != s2[i]:
                return s2[:i]
        return s1

LeetCode14最长公共前缀官方解决方案

LeetCode15：三数之和

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Notice that the solution set must not contain duplicate triplets.

代码出自team-learning-program/LeetCodeTencent/015 三数之和.md。解决方法的核心思想就是双指针，一个从前向后，一个从后向前。

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums = sorted(nums)

        result = []

        for i in range(0, len(nums) - 2):
            # 如果最小的数字大于0, 后面的操作已经没有意义
            if nums[i] > 0:
                break
            # 跳过三元组中第一个元素的重复数据
            if i > 0 and nums[i-1] == nums[i]:
                continue
            
            # 限制nums[i]是三元组中最小的元素
            l = i + 1
            r = len(nums) - 1            
            while l < r:
                sum = nums[i] + nums[l] + nums[r]
                if sum < 0:
                    l += 1
                elif sum > 0:
                    r -= 1
                else:
                    result.append([nums[i], nums[l], nums[r]])
                    # 跳过三元组中第二个元素的重复数据
                    while l < r and nums[l] == nums[l+1]:
                        l += 1
                    # 跳过三元组中第三个元素的重复数据
                    while l < r and nums[r] == nums[r-1]:
                        r -= 1                    
                    l += 1
                    r -= 1
        return result

LeetCode15三数之和官方解决方案

任务链接：

team-learning-program/LeetCodeTencent/011 盛最多水的容器.md
team-learning-program/LeetCodeTencent/014 最长公共前缀.md
team-learning-program/LeetCodeTencent/015 三数之和.md