• 算法之:练习题


    一、重排列

     方法一:

    class Solution:
        def isAnagram(self,s,t):
            """
            :param s: str
            :param t: str
            :return: bool
            """
            dict1 = {} #{'a':1,'b':2}
            dict2 = {}
            for ch in s:
                dict1[ch] = dict1.get(ch,0) + 1
            for ch in t:
                dict2[ch] = dict2.get(ch,0) + 1
            return dict1 == dict2
    

    方法二:

    class Solution:
        def isAnagram(self, s, t):
            """
            :param s: str
            :param t: str
            :return: bool
            """
            return sorted(list(s)) == sorted(list(t))

    二、二维二分查找

     

    方法一:

    class Solution:
        def searchMatrix(self, matrix, target):
            """
            :param matrix: List[List[int]]
            :param target: int
            :return: bool
            """
            #[[],[],[]]
            m = len(matrix) # 行数
            if m == 0:
                return False
            n = len(matrix[0])  #列数
            if n == 0:
                return False
            low = 0
            high = m * n - 1
    
            while low <= high:
                mid =(low + high) // 2
                x,y = divmod(mid,n)
    
                if matrix[x][y] > target:
                    high = mid - 1
                elif matrix[x][y] < target:
                    low = mid + 1
                else:
                    return True
            else:
                return  False

    方法二:

    class Solution:
        def searchMatrix(self, matrix, target):
            if not matrix or target is None:
                return False
    
            rows,cols = len(matrix), len(matrix[0])
            low, high = 0,rows * cols - 1
    
            while low <= high:
                mid = (low + high) / 2
                num = matrix[mid / cols][mid % cols]
    
                if num == target:
                    return  True
                elif num < target:
                    low = mid + 1
                else:
                    high = mid + 1
            return False

    三、 Two Sum

    1、题目描述

    2、方法一

    class Solution:
        def binary_search(self,li,left,right,val):
            while left <= right:   #候选区有值
                mid = (left + right) // 2
                if li[mid][0] == val:
                    return mid
                elif li[mid][0] > val:  #待查找的值在mid左侧
                    right = mid - 1
                else: #li[mid]< val 待查找的值在mid右侧
                    left = mid + 1
            else:
                return None
    def twoSum(self,nums,trarget):
        """
        :param self:
        :param nums: List[int]
        :param trarget: int
        :return:
        """
    
        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]
            b = trarget - a
            if b >= a:
            #print(b,i+1,len(nums)-1)
                j = self.binary_search(new_nums,i+1,len(nums)-1,b)
            else:
                j = self.binary_search(new_nums, 0, i - 1, b)
            if j:
                return sorted([new_nums[i][1],new_nums[j][1]])

    3、方法二

    def twoSum(self,nums,trarget):
        """
        :param self:
        :param nums: List[int]
        :param trarget: int
        :return: List[int]
        """
        #[1,4,6,7,9,3], 9
        #{1:0,4:1,6:2,7:3,9:4}
        d = {}
        for i in range(len(nums)):
            a = nums[i]
            b = trarget - a
            if b in d:
                return [d[b],i]
            else:
                d[a] = i 

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

    四、区域查询

     方法一:

    def find_a(nums,target):
        low = 0
        high = len(nums) - 1
        while low <= high:
            mid = (low + high) // 2
            if target <= nums[mid]:
                high = mid - 1
            else:
                low = mid + 1
    
        #[1,2,3,2,4,8,10]
    
        if low < len(nums):
            return low
        else:
            return  -1
    
    def find_b(nums,target):
        low = 0
        high = len(nums) - 1
        while low <= high:
            mid = (low + high) // 2
            if target < nums[mid]:
                high = mid - 1
            else:
                low = mid + 1
        if low < len(nums):
            return low
        else:
            return -1
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  • 原文地址:https://www.cnblogs.com/luoahong/p/9687353.html
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