代码随想训练营第七天（Python）｜ 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
关键点：减少复杂度判断 a+b 是否 等于 -(d+e).求和类的题目，利用好相反数。

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

383. 赎金信
1、字典

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        record = {}
        for i in magazine:
            record[i] = record.get(i, 0) + 1
        for j in ransomNote:
            if j not in record or record.get(j) == 0:
                return False
            else:
                record[j] = record.get(j) - 1
        return True

2、数组

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        record = [0]*26
        for i in ransomNote:
            record[ord(i)-ord("a")] += 1
        for j in magazine:
            record[ord(j)-ord("a")] -= 1
        for k in record:
            if k > 0:
                return False
        return True

15.三数之和
注意点：对于双指针需要添加到结果集后进行一个去重

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res = []
        n = len(nums)
        nums.sort()
        for i in range(n):
            if nums[i] > 0:
                return res
            if i > 0 and nums[i] == nums[i-1]: # 需要加入到结果集后再对 nums[i] 去重
                continue
            left, right = i+1, n-1
            while left < right:
                sum_val = nums[i] + nums[left] + nums[right]
                if sum_val < 0:
                    left += 1
                elif sum_val > 0:
                    right -= 1
                else:
                    res.append([nums[i], nums[left], nums[right]])
                    while left < right and nums[left] == nums[left+1]: # 对 nums[left] 去重
                        left += 1
                    while left < right and nums[right] == nums[right-1]: # 对 nums[right] 去重
                        right -= 1
                    left += 1
                    right -= 1
        return res

18. 四数之和
注意点：剪枝的位置。第二层的不能return res

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        res = []
        n = len(nums)
        nums.sort()
        for k in range(n):
            if nums[k] > target and target > 0: # 剪枝
                break
            if k > 0 and nums[k] == nums[k-1]:  #去重
                continue
            for i in range(k+1, n):
                if nums[i] + nums[k] > target and target > 0: # 剪枝
                    break
                if i > k+1 and nums[i] == nums[i-1]:  #去重
                    continue
                left, right = i+1, n-1
                while left < right:
                    sum_val = nums[k] + nums[i] + nums[left] + nums[right]
                    if sum_val < target:
                        left += 1
                    elif sum_val > target:
                        right -= 1
                    else:
                        res.append([nums[k], nums[i], nums[left], nums[right]])
                        while left < right and nums[left] == nums[left+1]:  # 去重
                            left += 1
                        while left < right and nums[right] == nums[right-1]:  # 去重
                            right -= 1
                        left += 1
                        right -= 1
        return res