Leetcode經典面試題 – 第1周

題目來源于Leetcode

報了一個百面機器學習面試的課程

每周都有定時打卡的作業

都是常出現于面試中的題

總結在此，時常溫習，

刷題小能手們覺得寫的不錯可以移步個人首頁

(ps:最近忙着筆試面試，更新太少)

1.雙指針（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.

給定一個遞增的數組，查找和等于目标值的兩個數

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

題目設計想用雙指針解決，雙指針複雜度O(n^2)，我這裡用字典解決這個問題，能減到O(n)的時間複雜度。具體就是，周遊目前的數組，查找目标值與目前周遊的數的差在不在數組中，在的話傳回索引即可

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.

在沒有排序的數組中找到第k大的數

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

題目讓我排序，還是偷懶，直接調用，然後切片索引

（能解決問題幹嘛繞路，但是假如初學還是老實從冒泡到快排都寫一遍吧，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.

給一個非空數組，傳回前k個頻繁出現的元素

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

題目讓我桶排序，桶排序指的是設定最小值到最大值的區間，指定n個桶，将待排序的數放入這些制定好的區間（桶）中，最後将區間中的數依次拿出即可

還是偷懶，用字典的值對應出現的次數，最後傳回即可

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.貪心（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.

發餅幹，每個孩子可能最少想要gi個餅幹，每塊餅幹j，尺寸為sj，如果sj>=gi，可以給那個孩子，孩子就滿足了。可以盡可能的滿足多少個孩子。

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

貪心問題，兩個數組升序排序，盡可能滿足條件的數量，根據貪心算法的解釋，也就是局部最優解，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