leetcode刷题● 454.四数相加II ● 383. 赎金信 ● 15. 三数之和 ● 18. 四数之和

454.四数相加II

四分二减小时间负杂度，由n**4变为n**2

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

 

383. 赎金信

字典法

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        d = {}
        for i in ransomNote:
            d[i] = d.get(i,0)+1
        for i in magazine:
            if i in d:
                d[i] = d.get(i) - 1
        for i in d:
            if d[i] > 0:
                return False
        return True

 5. 三数之和

注意去重循环很简单但去重很复杂很考验思维

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        result = []
        nums.sort()
        for i in range(len(nums)):
            if nums[i] > 0 :
                break  
            if i>0 and nums[i-1] == nums[i]:#a去重
                continue
            d = {}
            for j in range(i+1,len(nums)):
                c = 0 - nums[i] - nums[j]
                if j>i+2 and nums[j-1]==nums[j]==nums[j-2]:
                    continue
                if c in d :
                    result.append([nums[i],nums[j],c])
                    d.pop(c)
                else:
                    d[nums[j]] = j 

        return result


            

 18. 四数之和

与上题方法一样

也可以用字典法但时间复杂度较高很难在正规场合使用

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        d = {}
        ans = set()
        for i in range(len(nums)):
            d[nums[i]] = d.get(nums[i],0) + 1
        for i in range(len(nums)):
            for j in range(i+1,len(nums)):
                for k in range(j+1,len(nums)):
                     val = target - nums[i]- nums[j]-nums[k]
                     if val in d:
                         count = (nums[i]==val)+(nums[j]==val)+(nums[k]==val)
                         if d[val] > count:
                             ans.add(tuple(sorted([nums[i],nums[j],nums[k],val])))
        return [list(x) for x in ans]