哈希表（leetcode）2021.9.14

1、两数之和

1、遍历

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        n = len(nums)
        for i in range(n):
            for j in range(i+1,n):
                if target == nums[i] + nums[j]:
                    return [i,j]

2、哈希表

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        hash = dict()
        for i,num in enumerate(nums):
            if target-num in hash:
                return [hash[target-num],i]
            hash[nums[i]] = i

        return 

454、四数之和
1、哈希表
countAB = [’-1’:1,‘0’:3]

class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        countAB = collections.Counter(u + v for u in nums1 for v in nums2)
        ans = 0
        for i in nums3:
            for j in nums4:
                if -(i+j) in countAB:
                    ans += countAB[-(i+j)]
        return ans

560、和为 K 的子数组

class Solution(object):
    def subarraySum(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        fixSum = 0
        count = 0
        self.dict = dict()
        self.dict[0] = 1
        for num in nums:
            fixSum += num
            if fixSum-k in self.dict:
                count += self.dict[fixSum-k]
            if fixSum in self.dict:
                self.dict[fixSum] += 1
            else:
                self.dict[fixSum] = 1
        return count