leetcode刷题python之散数之和

首先是一个成功执行一次，但是提交却失败的程序。。。

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        result=[]
        length=len(nums)
        for i in range(len(nums))[1:length]:
            a=nums[i]
            left=0
            right=length-1
            # print(i)
            # print(nums)
            while left<i<right:
                num_one = nums[left]
                num_two = nums[i]
                num_three = nums[right]
                if (num_one+num_two+num_three == 0) and ([num_one,num_two,num_three] not in result):
                    result.append([num_one,num_two,num_three]) 
                    if left-i>1:left+=1
                elif num_one+num_two+num_three < 0:
                    left+=1
                else :
                    right-=1
                # print([num_one,num_two,num_three])
        return result

总算找了一个可以用的，不过真的是得做好多优化哎

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        result=[]
        length=len(nums)
        for i in range(length):
            if i > 0 and nums[i] == nums[i-1]:
                continue
            
            left=i+1
            right=length-1
            
            while left<right:
                # if nums[right]*nums[left]>0:break
                sum_num=nums[right]+nums[left]+nums[i]
                
                if (sum_num == 0) :
                    result_middle=[nums[right],nums[left],nums[i]]
                    result.append(result_middle) 
                    while left < right and nums[left] == nums[left+1]:
                        left += 1
                    while left < right and nums[right] == nums[right-1]:
                        right -= 1
                    left += 1
                    right -= 1

                    
                elif sum_num < 0:
                    left+=1
                else :
                    right-=1
                
        return result