Determine whether there are duplicate elements in the array

Array

principle

python How to use

Arrangement of elements in the list
sort Method

Given an array of integers , Determine whether there are duplicate elements .

If a value exists, it appears at least twice in the array , The function returns true . If every element in the array is different , Then return to false .
Their thinking

1. Sort existing arrays ,
2. Compare the elements in the array bit by bit

# The first version of the code ：
class Solution(object):
    def containsDuplicate(self, nums):
        """ :type nums: List[int] :rtype: bool """
        #1.  Sort the elements in the list 
        nums.sort()
        print "nums", nums
        n = len(nums)
        print n
        i = 0
        # while i <= n:
        # if nums[i] == nums[i+1]:
        # return True
        # else:
        # return False
        for i in range(n-1):
            if nums[i] == nums[i+1]:
                return True
            else:
                return False
            i = i + 1

The error message is as above ： Loop variable setting error , The value of the loop variable is always 0, The current two elements are the same , Can be returned as true, When there is only one element in the list , Unable to output ,
The modified code is as follows ：

class Solution(object):
    def containsDuplicate(self, nums):
        """ :type nums: List[int] :rtype: bool """
        nums.sort()
        n = len(nums)
        if n <=1:
            return False
        else: 
            for i in range(n-1):
                # print "i", i
                if nums[i] == nums[i+1]:
                    return True
            return False

The loop variable is set incorrectly ,i The value of is always 0, At this time, the element cannot be shifted for comparison , Only compare the first and second , The last two elements will not be compared

class Solution(object):
    def containsDuplicate(self, nums):
        """ :type nums: List[int] :rtype: bool """
        nums.sort()
        n = len(nums)
        for i in range(n-1):
            if nums[i] == nums[i+1]:
                return True
            i = i + 1     
        return False

Current The time complexity is ：O(nlogn), The space complexity is 1
Solution 2 ：
Hashtable

class Solution(object):
    def containsDuplicate(self, nums):
        """ :type nums: List[int] :rtype: bool """
        # Method 2 ： Hashtable 
        visted = set()
        for num in nums:
            if num in visted:
                return True
            visted.add(num)
        return False

Time complexity ：O(N), Spatial complexity O(N)
Personal opinion ： Space for time