1-15-18-454. 234Sum [Easy & Medium] Sum系列问题-优快云博客

博客主要围绕两数、三数、四数求和问题展开。对于 3Sum 和 4Sum 问题，时间复杂度为 O(n3)，需先排序，在循环和指针移动时处理重复值；4Sum II 问题时间复杂度为 O(n2)。

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1. Two Sum

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        dic = {}
        for i in range(len(nums)):
            if nums[i] in dic:
                return [dic[nums[i]], I]
            else:
                dic[target - nums[i]] = I

15. 3Sum

15.3Sum

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        nums.sort()
        m = len(nums)
        res = []
        for i in range(m-2):
            if i>0 and nums[i]==nums[i-1]:
                continue
            left = i + 1
            right = m - 1
            tgt = -nums[I]
            while left < right:
                if nums[left] + nums[right] == tgt:
                    res.append([nums[i], nums[left], nums[right]])
                    while left + 1 < right and nums[left+1] == nums[left]:
                        left += 1
                    while right - 1 > left and nums[right-1] == nums[right]:
                        right -= 1
                    left += 1
                    right -= 1
                elif nums[left] + nums[right] > tgt:
                    right -= 1
                else:
                    left += 1
        return res

18. 4Sum

O(n3)
首先要排序，重复出现在三个地方，第一层循环，第二层循环，和left，right指针，在第一层，第二层循环要记录上一个值，如果相同就continue，当满足条件，添加之后，需要将left和right指针都移动得到和之前不相同的位置上。

class Solution(object):
    def fourSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        res = []
        nums.sort()
        m = len(nums)
        pre_i = None
        for i in range(m-3):
            if nums[i] == pre_i:
                continue
            pre_i = nums[i]
            pre_j = None
            for j in range(i+1, m-2):
                if nums[j] == pre_j:
                    continue
                pre_j = nums[j]
                tgt = target - nums[i] - nums[j]
                left = j + 1
                right = m - 1
                while left < right:
                    if nums[left] + nums[right] == tgt:
                        res.append([nums[i],nums[j],nums[left],nums[right]])
                        while left + 1 < right and nums[left+1] == nums[left]:
                            left += 1
                        while right - 1 > left and nums[right-1] == nums[right]:
                            right -= 1
                        left += 1
                        right -= 1
                    elif nums[left] + nums[right] > tgt:
                        right -= 1
                    else:
                        left += 1
        return res

454. 4Sum II

O(n2)

class Solution(object):
    def fourSumCount(self, A, B, C, D):
        """
        :type A: List[int]
        :type B: List[int]
        :type C: List[int]
        :type D: List[int]
        :rtype: int
        """
        res = 0
        mydic = {}
        m, n = len(A), len(B)
        for i in range(m):
            for j in range(n):
                temp = 0 - A[i] - B[j]
                mydic[temp] = mydic.get(temp,0) + 1
        m, n = len(C), len(D)
        for i in range(m):
            for j in range(n):
                temp = C[i] + D[j]
                if temp in mydic:
                    res += mydic[temp]
        return res