代码随想录算法公开课day5

242. 有效的字母异位词 - 力扣（LeetCode）

# map
class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False
        hash = {}
        for i in s:
            hash[i] = hash.get(i,0) + 1
        for i in t:
            hash[i] = hash.get(i,0) - 1

        for key,value in hash.items():
            if value != 0:
                return False

        return True

数组
class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False
        hash = [0] * 26
        for i in s:
            hash[ord(i) - ord('a')] += 1
        for i in t:
            hash[ord(i) - ord('a')] -= 1
        for i in hash:
            if i !=0 :
                return False
        return True

349. 两个数组的交集 - 力扣（LeetCode）

# 字典和集合
class Solution:
    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
        hash = {}
        for i in nums1:
            hash[i] = hash.get(i,0)
        result = []
        for i in nums2:
            if i in hash.keys():
                result.append(i)
                del hash[i]
        return result

# 数组
class Solution:
    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
        hash1 = [0] * 1001
        hash2 = [0] * 1001
        for i in nums1:
            hash1[i] += 1
        for i in nums2:
            hash2[i] += 1
        result = []
        for i in range(1001):
            if hash1[i] * hash2[i] != 0:
                result.append(i)
        return result

# 集合
class Solution:
    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
        return list(set(nums1) & set(nums2))

1. 两数之和 - 力扣（LeetCode）

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

454. 四数相加 II - 力扣（LeetCode）

class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        hashd = {}
        for i in nums1:
            for j in nums2:
                hashd[i + j] = hashd.get(i+j,0) + 1
        result = 0
        for i in nums3:
            for j in nums4:
                if -(i + j) in hashd:
                    result += hashd[-(i+j)]
        return result

15. 三数之和 - 力扣（LeetCode）

# 双指针
class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        lenth = len(nums)
        result = []
        if nums[0] > 0:
            return result
        for a in range(lenth -1):
            if a != 0 and nums[a] == nums[a-1]:
                continue
            left = a + 1
            right = lenth - 1
            while left < right:
                _sum = nums[left] + nums[right] + nums[a]
                if  _sum == 0:
                    result.append([nums[a],nums[left],nums[right]])
                    while left < lenth-2 and nums[left] == nums[left+1]:
                        left += 1
                    while right > a+2 and nums[right] == nums[right-1]:
                        right -= 1
                    left += 1
                    right -= 1
                elif _sum > 0:
                    right -= 1
                else:
                    left += 1

        return result