查找表_leetcode15

#coding=utf-8
# 解题思路：查找表 n2 20190302 找工作期间
# 排序数组的双指针求和法 : (0,n) -> (a,b)
#  其中a只能增加  b只能减少 : 搜索方向的问题 ,不然会到重复的状态

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
#  时间复杂度 n3
        if len(nums) < 3:
            return []
        result = []
        for i in range(len(nums) - 1):
            for j in range(i + 1, len(nums)):
                sum = nums[i] + nums[j]
                if -sum in nums:
                    k = nums.index(-sum)
                    if k != i and k != j:
                        temp = sorted([nums[i], nums[j], nums[k]])
                        if temp not in result:
                            result.append(temp)
        return result



class Solution2(object):
    def threeSum(self, nums):
        if len(nums) < 3:
            return []
        result = []
        nums = sorted(nums)
        for index, item in enumerate(nums):
            target = -item
            start = index + 1
            end = len(nums) - 1
            if item > 0 or nums[end] < 0:
                break
            if index > 0 and item == nums[index-1]:
                continue
            while start < end:
                if nums[end] < 0:
                    break
                if nums[start] + nums[end] == target:
                    temp_list = [nums[index], nums[start], nums[end]]
                    result.append(temp_list)

                    while start < end and nums[start] == nums[start+1]:
                        start = start + 1
                    while start < end and nums[end] == nums[end-1]:
                        end = end - 1

                    start = start + 1
                    end = end - 1

                elif nums[start] + nums[end] < target:
                    start = start + 1
                else:
                    end = end - 1

        return result