leetcode_15 3Sum(TwoPointers)

Posted 2020-09-18

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]
]

---

易错：出现重复的结果

暴力枚举三个数复杂度为O(N^3)。先考虑2Sum的做法，假设升序数列a，对于一组解ai,aj, 另一组解ak,al ，必然满足 i<k j>l 或 i>k j<l, 因此我们可以用两个指针，初始时指向数列两端
指向数之和大于目标值时，右指针向左移使得总和减小，反之左指针向右移。由此可以用O(N)的复杂度解决2Sum问题，3Sum则枚举第一个数O(N^2)，使用有序数列的好处是，在枚举和移动指针时值相等的数可以跳过，省去去重部分

Python实现：

class Solution(object):
    def threeSum(self, nums):
        nums.sort()
        ans = []  # 结果
        for i in range(0, len(nums)-2):
            if i and nums[i] == nums[i-1]:
                continue
            target = -1*nums[i]
            left = i + 1
            right = len(nums)-1
            while left < right:
                if nums[left] + nums[right] == target:
                    ans.append([nums[i], nums[left], nums[right]])
                    left += 1
                    right -= 1
                    while left < right and nums[left] == nums[left-1]:  #避免结果重复 ！！！
                        left += 1
                    while left < right and nums[right] == nums[right+1]:
                        right -= 1
                elif nums[left] + nums[right] < target:
                    left += 1
                    # while left < right and nums[left] == nums[left+1]:
                    #     left += 1
                else:
                    right -= 1
                    # while left < right and nums[right] == nums[right-1]:
                    #     right -= 1
        return ans