算法——查找排序相关面试题和leetcode使用

时间:2022-11-30 14:17:35

1、给两个字符串s和t,判断t是否为s的重新排列后组成的单词。

  • s = "anagram", t = "nagaram", return true.
  • s = "rat", t = "car", return false.
  • leetcode地址:https://leetcode.com/problems/valid-anagram/description/

(1)解法一:排序,O(n*logn)

class Solution:
    def isAnagram(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        ss = list(s)
        tt = list(t)
        ss.sort()
        tt.sort()
        return ss == tt
"""
输入:"anagram"、"nagaram"
输出:true
Runtime: 32 ms
"""

  排序方法简写如下:

class Solution:
    def isAnagram(self, s, t):
        return sorted(list(s)) == sorted(list(t))

(2)解法二:判断字母数量一致,时间复杂度O(n)

class Solution:
    def isAnagram(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        dict1 = {}   # 用字典来维护字符的数量
        dict2 = {}
        for ch in s:
            dict1[ch] = dict1.get(ch, 0) + 1   # 没有就新建,有就加1
        for ch in t:
            dict2[ch] = dict2.get(ch, 0) + 1
        return dict1 == dict2

"""
输入:"anagram","nagaram"
输出:true
Runtime: 32 ms
"""

2、给定一个m*n的二维列表,查找一个数是否存在。

  列表有下列特性:

  • 每一行的列表从左到右已经排序好。
  • 每一行第一个数比上一行最后一个数大。
  • leetcode地址:https://leetcode.com/problems/search-a-2d-matrix/description/

  算法——查找排序相关面试题和leetcode使用

(1)解法一:线性查找查找,O(mn)

class Solution:
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        for line in matrix:
            if target in line:
                return True
        return False

(2)解法二:二分查找O(logn)

class Solution:
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        h = len(matrix)  # 高
        if h == 0:
            return False
        
        w = len(matrix[0])  # 列
        if w == 0:
            return False
        
        left = 0
        right = w * h - 1
        
        while left  <= right:
            mid = ((left + right)) // 2
            i = mid // w
            j = mid % w   
            if matrix[i][j] == target:
                return True
            elif matrix[i][j] > target:
                right = mid - 1
            else:
                left = mid + 1
        else:
            return False

3、给定一个列表和一个整数,设计算法找到两个数的下标,使得两个数之和为给定的整数。

  保证肯定仅有一个结果。

  leetcode地址:https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/

  例如,列表[1,2,5,4]与目标整数3,1+2=3,结果为(0,1).

(1)方法一:通过二分查找,找到需要的数字。时间复杂度:O(nlogn)

  首先确定第一个数,再通过给定整数确定要查找的数,通过二分查找到需要的数。

class Solution:
    def binary_search(self, li, left, right,val):
        """
        二分查找
        :param li: 输入的列表
        :param val: 输入的待查找的值
        :return:
        """
        while left <= right:  # 说明候选区有值
            mid = (left + right) // 2   # 因为是下标, 因此要整除2
            if li[mid] == val:
                # 找到待查找的值返回index
                return mid
            elif li[mid] > val:
                # 待查找的值在mid左侧
                right = mid - 1   # 更新候选区
            else:  # li[mid] < val
                # 待查找的值在mid右侧
                left = mid + 1    # 更新候选区
        else:
            # 没有找到
            return None
        
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for i in range(len(nums)):
            a = nums[i]
            b = target - a
            if b >= a:
                j = self.binary_search(nums, i+1, len(nums)-1, b)
            else:
                j = self.binary_search(nums, 0, i-1, b)
            if j:
                break
        return sorted([i+1,j+1])
        

(2)方法二:针对已经排好序的列表

  leetcode地址:https://leetcode.com/problems/two-sum/description/

class Solution:
    def binary_search(self, li, left, right,val):
        """
        二分查找
        :param li: 输入的列表
        :param val: 输入的待查找的值
        :return:
        """
        while left <= right:  # 说明候选区有值
            mid = (left + right) // 2   # 因为是下标, 因此要整除2
            if li[mid][0] == val:
                # 找到待查找的值返回index
                return mid
            elif li[mid][0] > val:
                # 待查找的值在mid左侧
                right = mid - 1   # 更新候选区
            else:  # li[mid] < val
                # 待查找的值在mid右侧
                left = mid + 1    # 更新候选区
        else:
            # 没有找到
            return None
        
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        new_nums = [[num, i] for i,num in enumerate(nums)]
        new_nums.sort(key=lambda x:x[0])
        
        for i in range(len(new_nums)):
            a = new_nums[i][0]   # 数
            b = target - a
            if b >= a:
                j = self.binary_search(new_nums, i+1, len(new_nums)-1, b)
            else:
                j = self.binary_search(new_nums, 0, i-1, b)
            if j:
                break
        return sorted([new_nums[i][1], new_nums[j][1]])