【LeetCode】15. 3Sum(Python)

Problem

Algorithmic thinking

讨论解决方案 https://discuss.leetcode.com/topic/75883/python-solution-with-detailed-explanation

3Sum https://leetcode.com/problems/3sum/

基于排序的算法

• a + b = -c。3SUM减少到2SUM问题。

处理2SUM中的重复项

• 假设索引s和e正在排序数组中形成解决方案。现在给nums [s]，有一个独特的nums [e]，这样nums [s] + nums [e] = Target。因此，如果nums [s + 1]与nums [s]相同，那么在范围s + 1到e中搜索将给出一个重复的解决方案。因此，我们必须将s移动到nums [s]！= nums [s-1]以避免重复。

                        while s<e and nums[s] == nums[s-1]:
                            s = s+1

处理3SUM中的重复项

• 想象一下，我们在索引i处，我们已经从索引i + 1调用了2SUM问题到数组的末尾。现在一旦2SUM终止，我们将有一个包含nums [i]的所有三元组的列表。为了避免重复，我们必须跳过所有nums [i]，其中nums [i] == nums [i-1]。

            if i > 0 and nums[i] == nums[i-1]:
                continue        

Python解法：

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        nums.sort()
        N, result = len(nums), []
        for i in range(N):
            if i > 0 and nums[i] == nums[i-1]:
                continue
            target = nums[i]*-1
            s,e = i+1, N-1
            while s<e:
                if nums[s]+nums[e] == target:
                    result.append([nums[i], nums[s], nums[e]])
                    s = s+1
                    while s<e and nums[s] == nums[s-1]:
                        s = s+1
                elif nums[s] + nums[e] < target:
                    s = s+1
                else:
                    e = e-1
        return result

相关文档：
https://leetcode.com/problems/3sum/discuss/7498/Python-solution-with-detailed-explanation

Python3 other solution

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
3Sum 三数之和
eg: -1+0+1=0

"""


class Solution:

    def threeSum(self, nums):
        """
        @param nums: list[int]
        @return: List[List[int]]
        """
        d = {}
        for val in nums:
            d[val] = d.get(val, 0) + 1

        pos = [x for x in d if x > 0]
        neg = [x for x in d if x < 0]

        res = []
        if d.get(0, 0) > 2:
            res.append([0, 0, 0])

        for x in pos:
            for y in neg:
                s = -(x + y)
                if s in d:
                    if s == x and d[x] > 1:
                        res.append([x, x, y])
                    elif s == y and d[y] > 1:
                        res.append([x, y, y])
                    elif y < s < x:
                        res.append([x, y, s])
        return res


if __name__ == '__main__':
    sl_1 = [-1, 0, 1, -2, 2, -3, 3]
    result = Solution().threeSum(sl_1)
    print(result)