class Solution:def threeSumClosest(self, nums, target):new_nums = 0 # 创建储存结果的变量nums.sort()  # 对原有数组进行排序处理minaway = 100for i in range(len(nums)-2): # 从小到大遍历所有的元素 left  = i+1  # 创建左指针right = len(nums)-1  # 创建右指针while left < right:ident = nums[left] + nums[right] + nums[i]  #将nums[i]加上最大的 和 最小的 数字away  = abs(ident - target) # 计算三数之和与target的距离if away < minaway : # 判断距离是否为已知的最近距离minaway = away  # 对最小距离指针进行更新 new_nums = ident #对结果进行更新if new_nums < target :left += 1  # 更新左指针elif new_nums > target :right -= 1 #更新右指针else : #当数据全部遍历完毕return new_nums  # 返回结果

class Solution:def threeSumClosest(self, nums, target):minaway = 100000000000away = 0for i in range(len(nums)):for j in range(i+1,len(nums)):for k in range(j+1,len(nums)):if abs(nums[i]+nums[j]+nums[k]-target) < minaway:away = nums[i]+nums[j]+nums[k]minaway = abs(nums[i]+nums[j]+nums[k]-target)return away

class Solution:def removeDuplicates(self, nums):""":type nums: List[int]pe: int"""i=0while i<len(nums)-1:if nums[i]==nums[i+1]:nums.remove(nums[i])else:i=i+1return len(nums)

class Solution:def removeElement(self, nums, val):while val in nums:nums.remove(val)return len(nums)

class Solution:def nextPermutation(self, nums):""":type nums: List[int]:rtype: void Do not return anything, modify nums in-place instead."""i = len(nums)-1while i>0:if nums[i] > nums[i-1]:t = nums[i-1]nums[i-1] = nums[i]nums[i] = tbreaki -= 1else:nums = nums[::-1]

class Solution:def nextPermutation(self, nums):""":type nums: List[int]:rtype: void Do not return anything, modify nums in-place instead."""i = len(nums) - 1while i>0:if nums[i] <= nums[i-1]:  # 从末至首寻找升序子列表i -= 1else :  # 如果升序子列表中断，则跳出循环breakif i > 0:  # 如果i>0，则证明num列表不全是升序元素j = len(nums) - 1while nums[j] <= nums[i-1] : # 从尾部开始找第一个比断点大的数字确保是最优解j -= 1nums[j],nums[i-1] = nums[i-1],nums[j]  # 断点与该数字进行值的互换new_list = sorted(nums[i:])  # 对断点之后进行升序排列，确保数字最小k = 0for m in range(i, len(nums)):nums[m] = new_list[k]k += 1else :nums.sort()