Leetcode刷题笔记1 哈希表part02

leetcode454 四数相加 ||

 字典法

  先计算n1 + n2，将其和制成一个字典；然后计算-n3-n4是否在字典中。

class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        hashmap = dict()
        for n1 in nums1:
            for n2 in nums2:
                if n1 + n2 in hashmap:
                    hashmap[n1+n2] += 1
                else:
                    hashmap[n1+n2] = 1
        count = 0
        for n3 in nums3:
            for n4 in nums4:
                key = -n3 - n4 
                if key in hashmap:
                    count += hashmap[key]
        return count

leetcode 383 赎金信

collections库方法（Counter）

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

哈希表法

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        if len(ransomNote) > len(magazine):
            return False
        ransom_count = [0] * 26
        magazine_count = [0] * 26
        for c in ransomNote:
            ransom_count[ord(c) - ord('a')] += 1
        for c in magazine:
            magazine_count[ord(c) - ord('a')] += 1
        return all(ransom_count[i] <= magazine_count[i] for i in range(26))

leetcode15 三数之和

初步思路是通过哈希表法确定第三个数，但是题目中的不重复无法满足

本题的指导思路首先是双指针法，其次在去重的时候代码应该是‘if i > 0 and nums[i] == nums[i - 1]:’，因为我们需要去重的是相同的三元组而不是组内相同的数。

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        n = len(nums)
        ans = []
        if n < 3:
            return ans
        nums.sort()
        for i in range(n):
            if nums[0] > 0:
                return ans
            if i > 0 and nums[i] == nums[i-1]:
                continue
            l,r = i+1,n-1
            while(l<r):
                sum = nums[i] + nums[l] + nums[r]
                if sum == 0:
                    ans.append([nums[i],nums[l],nums[r]])
                    while l < r and nums[r] == nums[r-1]:
                        r -= 1
                    while l < r and nums[l] == nums[l+1]:
                        l += 1
                    l += 1
                    r -= 1
                elif sum > 0:
                    r -= 1
                else:
                    l += 1
        return ans

leetcode18 四数之和

四数之和和三数之和有些相似，不同的是需要再三数之和外围加一个for循环。同时由于四数之和中要求四数不重复，所以需要再循环前后加边界判断

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        n = len(nums)
        ans = []
        nums.sort()
        if n < 4 or not nums:
            return ans
        for i in range(n - 3):
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            for j in range(i + 1, n):
                if j > i + 1 and nums[j] == nums[j - 1]:
                    continue
                l, r  = j + 1, n - 1
                while l < r:
                    sum = nums[i] + nums[j] + nums[l] + nums[r]
                    if sum == target:
                        ans.append([nums[i],nums[j],nums[l],nums[r]])
                        while l < r and nums[l] == nums[l+1]:
                            l += 1
                        while l < r and nums[r] == nums[r-1]:
                            r -= 1
                        l += 1
                        r -= 1
                    elif sum > 0:
                        r -= 1
                    else:
                        l += 1
        return ans