Leetcode经典面试题 -- 第1周

Leetcode经典面试题 – 第1周

题目来源于Leetcode
报了一个百面机器学习面试的课程
每周都有定时打卡的作业
都是常出现于面试中的题
总结在此，时常温习，
刷题小能手们觉得写的不错可以移步个人主页
(ps:最近忙着笔试面试，更新太少)

1.双指针（Leetcode 167）

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

给定一个递增的数组，查找和等于目标值的两个数

Example:
Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.

思路

题目设计想用双指针解决，双指针复杂度O(n^2)，我这里用字典解决这个问题，能减到O(n)的时间复杂度。具体就是，遍历当前的数组，查找目标值与当前遍历的数的差在不在数组中，在的话返回索引即可

class Solution:
    def twoSum(self, nums, target):
        dict = {}
        first = -1
        second = -1
        for i in range(len(nums)):
            complement = target - nums[i]
            if complement in dict:
                first = dict[complement]
                second = i
                break
            dict[nums[i]] = i
        return first+1, second+1

2.排序（Leetcode 215）

Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.

在没有排序的数组中找到第k大的数

Example 1:
Input: [3,2,1,5,6,4] and k = 2
Output: 5

Example 2:
Input: [3,2,3,1,2,4,5,5,6] and k = 4
Output: 4

思路

题目让我排序，还是偷懒，直接调用，然后切片索引
（能解决问题干嘛绕路，但是假如初学还是老实从冒泡到快排都写一遍吧，嘿嘿）

class Solution(object):
    def findKthLargest(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        nums.sort()
        return nums[-k]

3.桶排序（Leetcode 347）

Given a non-empty array of integers, return the k most frequent elements.

给一个非空数组，返回前k个频繁出现的元素

Example 1:
Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:
Input: nums = [1], k = 1
Output: [1]

思路

题目让我桶排序，桶排序指的是设定最小值到最大值的区间，指定n个桶，将待排序的数放入这些制定好的区间（桶）中，最后将区间中的数依次拿出即可
还是偷懒，用字典的值对应出现的次数，最后返回即可

class Solution(object):
    def topKFrequent(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        d = {}
        for i in range(len(nums)):
            d[nums[i]] = d.get(nums[i], 0) + 1
        res = []
        for i in range(k):
            res.append(max(d, key=d.get))
            d[res[i]] = 0
        return res
                    

4.贪心（Leetcode 455）

Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie. Each child i has a greed factor gi, which is the minimum size of a cookie that the child will be content with; and each cookie j has a size sj. If sj >= gi, we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.

发饼干，每个孩子可能最少想要gi个饼干，每块饼干j，尺寸为sj，如果sj>=gi，可以给那个孩子，孩子就满足了。可以尽可能的满足多少个孩子。

Example 1:
Input: [1,2,3], [1,1]
Output: 1
Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3.
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1.

Example 2:
Input: [1,2], [1,2,3]
Output: 2
Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2.
You have 3 cookies and their sizes are big enough to gratify all of the children,
You need to output 2.

思路

贪心问题，两个数组升序排序，尽可能满足条件的数量，根据贪心算法的解释，也就是局部最优解，它就是答案

class Solution(object):
    def findContentChildren(self, g, s):
        g.sort()
        s.sort()
        i, j = 0, 0
        while i < len(g) and j < len(s):
            if s[j] >= g[i]:
                i += 1
            j += 1
        return i