力扣--260只出现一次的数字I-III

题目I

Python 代码

1 利用字典计数

def singleNumber(self,nums):
    dict={
     }
    for i in nums:
        if i not in dict:
            dict[i]=1
        else:
            dict[i]+=1
    for key, value in dict.items():
        if value==1:
            return key
    return 0

2 排序比较判断

def singleNumber(self,nums):
    nums.sort()
    length = len(nums)
    if length < 3:
        return nums[0]
    i = 0
    while i < length - 2 :
        if nums[i] != nums[i+1]:
            return nums[i]
        else:
            i += 2
    return nums[-1]

3 集合交叉法

def singleNumber(self, nums):        
    nums.sort()        
    n = list(set(nums[::2]) - set(nums[1::2]))[0]        
    return n

题目II

python代码

def singleNumber(self,nums):
    dict={
     }
    for i in nums:
        if i not in dict:
            dict[i]=1
        else:
            dict[i]+=1
    for key, value in dict.items():
        if value==1:
            return key
    return 0

题目III

python代码

def singleNumber(self,nums):
    dict={
     }
    nums_new=[]
    for i in nums:
        if i not in dict:
            dict[i]=1
        else:
            dict[i]+=1
    for key, value in dict.items():
        if value==1:
            nums_new.append(key)
    return nums_new