leetcode(26)哈希表系列题目

242. 有效的字母异位词

哈希表方法，可适应更大规模字符集

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        dict_ = {}
        for ch in s:
            if ch in dict_:
                dict_[ch] += 1
            else:
                dict_[ch] = 1
        for ch in t:
            if ch in dict_:
                dict_[ch] -= 1
            else:
                dict_[ch] = 1
        for v in dict_.values():
            if v != 0:
                return False
        return True

或者取巧用Counter()函数

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        dict_s, dict_t = Counter(s), Counter(t)
        if dict_s == dict_t:
            return True
        else:
            return False

349. 两个数组的交集

注意：if hash.get(n):如果键n对应的值为0则返回False；而if n in hash:会返回True

class Solution:
    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
        res = []
        hash = {}
        for n in nums1:
            if n not in hash:
                hash[n] = 1
        for n in nums2:
            # if n in hash:
            if hash.get(n):
                hash[n] = 0
                res.append(n)
        return res

取巧的做法：

class Solution:
    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
        res = []
        set1, set2 = set(nums1), set(nums2)
        for num in set1:
            if num in set2:
                res.append(num)
        return res

1. 两数之和

用枚举更方便，就不需要通过索引再去取当前位置的值

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        records = dict()
        for idx, val in enumerate(nums):
            if target - val in records:
                return [records[target - val], idx]
            else:
                records[val] = idx

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        dict_ = {}
        for i in range(len(nums)):
            if target - nums[i] in dict_:
                return [dict_[target - nums[i]], i]
            else:
                dict_[nums[i]] = i

454. 四数相加 II

把num1和num2看作一个整体，num3和num4看作一个整体

class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        res = 0
        records = dict()
        for i in nums1:
            for j in nums2:
                if i + j not in records:
                    records[i + j] = 1
                else:
                    records[i + j] += 1
        for i in nums3:
            for j in nums4:
                if -(i + j) in records:
                    res += records[-(i + j)]
        return res

383. 赎金信

Counter()函数可以直接相减
若ransomNote中的字符在magazine没有出现则会返回Counter({'a': 1})
否则返回Counter()，即为空的

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        if len(ransomNote) > len(magazine):
            return False
        return not Counter(ransomNote) - Counter(magazine)

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        recordsR, recordsM = Counter(ransomNote), Counter(magazine)
        for s in recordsR:
            if recordsR[s] > recordsM[s]:
                return False
        return True

705. 设计哈希集合

不符合要求的做法：

class MyHashSet:

    def __init__(self):
        self.hashset = set()

    def add(self, key: int) -> None:
        self.hashset.add(key)

    def remove(self, key: int) -> None:
        if key in self.hashset:
            self.hashset.remove(key)

    def contains(self, key: int) -> bool:
        if key in self.hashset:
            return True
        return False

规范题解见：详解 HashSet 的设计：在时间和空间上做权衡

706. 设计哈希映射

不符合要求的做法：

class MyHashMap:

    def __init__(self):
        self.hashmap = dict()

    def put(self, key: int, value: int) -> None:
        self.hashmap[key] = value

    def get(self, key: int) -> int:
        if key in self.hashmap:
            return self.hashmap[key]
        return -1

    def remove(self, key: int) -> None:
        if key in self.hashmap:
            del self.hashmap[key]

规范题解见：详解 HashMap 的设计：在时间和空间上做权衡