Leetcode 3670. Maximum Product of Two Integers With No Common Bits

• Leetcode 3670. Maximum Product of Two Integers With No Common Bits
  • 1. 解题思路
  • 2. 代码实现

• 题目链接：3670. Maximum Product of Two Integers With No Common Bits

1. 解题思路

这一题我的思路挺暴力的，就是先构造一个trie树，然后搜索对任意$n$的最大的补数，然后考察所有结果的最大值。

然后直接的实现出现了超时的问题，因此我做了不少剪枝修补，感觉不是最好的实现方法，有兴趣的读者建议还是再想想有没有什么更好的解法吧……

2. 代码实现

给出python代码实现如下：

DIGIT_NUM = 20

class Trie:
    def __init__(self):
        self.trie = {}
    
    def add(self, num):
        word = bin(num)[2:].rjust(DIGIT_NUM, "0")
        trie = self.trie
        for c in word:
            trie = trie.setdefault(c, {})
        trie["eos"] = num

    @lru_cache(None)
    def find(self, num):
        word = bin(num)[2:].rjust(DIGIT_NUM, "0")
        trie = self.trie

        def dfs(idx, trie):
            if idx == DIGIT_NUM:
                return trie.get("eos", 0)
            if trie == {}:
                return 0
            if word[idx] == "1":
                return dfs(idx+1, trie["0"]) if "0" in trie else 0
            else:
                return max(dfs(idx+1, trie.get("0", {})), dfs(idx+1, trie.get("1", {})))
        
        return dfs(0, trie)

@lru_cache(None)
def get_rev(num):
    num = bin(num)[2:]
    digits = [1-int(x) for x in num]
    ans = 0
    for d in digits:
        ans = 2*ans+d
    return ans

class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        snums = set(nums)
        nums = list(snums)
        trie = Trie()
        for num in nums:
            trie.add(num)

        nums = sorted(nums, reverse=True)
        ans = 0
        for num in nums:
            rev = get_rev(num)
            if num * num <= ans:
                break
            if get_rev(num) in snums:
                ans = max(ans, num * get_rev(num))
                continue
            ans = max(ans, num * trie.find(num))
        return ans

提交代码评测得到：耗时16912ms，占用内存319.43MB。