[LeetCode] 15. 3Sum 三数之和 @python

Description

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

---

给定一个包含 n 个整数的数组 S，是否存在属于 S 的三个元素 a，b，c 使得 a + b + c = 0 ？找出所有不重复的三个元素组合使三个数的和为零。

注意：结果不能包括重复的三个数的组合。

例如, 给定数组 S = [-1, 0, 1, 2, -1, -4]，

一个结果集合为：
[
  [-1, 0, 1],
  [-1, -1, 2]
]

---

Solutions

需要求的就是a+b=(-c)，然后在余下的链表里做两数相加即可。注意去重！

# -*- coding: utf-8 -*-
"""
Created on Sat Mar 17 16:30:25 2018

@author: Saul
"""
class Solution:
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        ans = []
        nums.sort()
        for i in range(len(nums)-2):
            if i == 0 or nums[i] > nums[i-1]:
                left = i+1
                right = len(nums)-1
                while left < right:
                    ident = nums[left] + nums[right] + nums[i]
                    if ident == 0:
                        ans.append([nums[i], nums[left], nums[right]])
                        left += 1; right -= 1
                        while left < right and nums[left] == nums[left-1]:    # skip duplicates
                            left += 1
                        while left < right and nums[right] == nums[right+1]:
                            right -= 1
                    elif ident < 0:
                        left += 1
                    else:
                        right -= 1
        return ans