Leetcode刷题笔记7：哈希表2

454.四数之和2

要善用哈希表和Counter，时间复杂度O(n2)。

class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        cnt = Counter([x+y for x in nums1 for y in nums2])
        return sum(cnt[-x-y] for x in nums3 for y in nums4)

383 赎金信

简单题，秒解～

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        cnt = Counter(list(magazine))
        for j in list(ransomNote):
            cnt[j] -= 1
        for val in cnt:
            if cnt[val] < 0:
                return False
        return True   

15.三数之和

看了题解自己写的。这道题是热门题，要及时复习！！注意边界条件。
看这道题之前，先看一下【167.两数之和2】这道题，排序后做法一样，
注意：要跳过重复的数字。不能先通过set操作，因为比如[-1,-1,2]这种情况下，确实某些数字要被使用2次。

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums =  sorted((nums))
        res = []
        for i in range(len(nums)-2):
            if i >0 and nums[i] ==nums[i-1]:
                continue
            j = i + 1
            k = len(nums) - 1
            while j < k:
                s = nums[i] + nums[j] + nums[k]
                if s == 0:
                    res.append([nums[i],nums[j],nums[k]])
                    while (j+1)<=(len(nums)-2) and nums[j] == nums[j+1]:
                        j+= 1
                    j += 1
                    while k-1>=0 and nums[k] == nums[k-1]:
                        k-=1
                    k -= 1
                elif s > 0:
                    k -= 1
                else:
                    j += 1
        return res

18.四数之和

18.四数之和：三数之和的升级版，在三数之和外面套一个循环，其他不变

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        nums.sort()
        n = len(nums)
        res = []
        for i in range(n-3):
            if i>0 and nums[i] == nums[i-1]:
                continue
            for j in range(i+1,n-2):
                if j>i+1 and nums[j] == nums[j-1]:
                    continue
                k = j + 1
                l = n - 1
                while k < l:
                    s = nums[i] + nums[j] + nums[k] + nums[l]
                    if s > target:
                        l -= 1
                    elif s < target:
                        k += 1
                    else:
                        res.append([nums[i],nums[j],nums[k],nums[l]])
                        k += 1
                        while k <l and nums[k] == nums[k-1]:
                            k+=1
                        l -= 1
                        while k<l and nums[l] == nums[l+1]:
                            l -= 1
        return res