leetcode 432. All O`one Data Structure

谁践踏了优雅 2022-06-04 10:20 165阅读 0赞

Implement a data structure supporting the following operations:

Inc(Key) - Inserts a new key with value 1. Or increments an existing key by 1. Key is guaranteed to be a non-empty string.  
Dec(Key) - If Key’s value is 1, remove it from the data structure. Otherwise decrements an existing key by 1. If the key does not exist, this function does nothing. Key is guaranteed to be a non-empty string.  
GetMaxKey() - Returns one of the keys with maximal value. If no element exists, return an empty string “”.  
GetMinKey() - Returns one of the keys with minimal value. If no element exists, return an empty string “”.  
Challenge: Perform all these in O(1) time complexity.

不喜欢这样的题，这是网上参考的答案，就这么做吧

代码如下：

class AllOne {
    public:
    
        void inc(string key) {
    
            // If the key doesn't exist, insert it with value 0.
            if (!bucketOfKey.count(key))
                bucketOfKey[key] = buckets.insert(buckets.begin(), {
       0, {key}});
    
            // Insert the key in next bucket and update the lookup.
            auto next = bucketOfKey[key], bucket = next++;
            if (next == buckets.end() || next->value > bucket->value + 1)
                next = buckets.insert(next, {bucket->value + 1, {}});
            next->keys.insert(key);
            bucketOfKey[key] = next;
    
            // Remove the key from its old bucket.
            bucket->keys.erase(key);
            if (bucket->keys.empty())
                buckets.erase(bucket);
        }
    
        void dec(string key) {
    
            // If the key doesn't exist, just leave.
            if (!bucketOfKey.count(key))
                return;
    
            // Maybe insert the key in previous bucket and update the lookup.
            auto prev = bucketOfKey[key], bucket = prev--;
            bucketOfKey.erase(key);
            if (bucket->value > 1) {
                if (bucket == buckets.begin() || prev->value < bucket->value - 1)
                    prev = buckets.insert(bucket, {bucket->value - 1, {}});
                prev->keys.insert(key);
                bucketOfKey[key] = prev;
            }
    
            // Remove the key from its old bucket.
            bucket->keys.erase(key);
            if (bucket->keys.empty())
                buckets.erase(bucket);
        }
    
        string getMaxKey() {
            return buckets.empty() ? "" : *(buckets.rbegin()->keys.begin());
        }
    
        string getMinKey() {
            return buckets.empty() ? "" : *(buckets.begin()->keys.begin());
        }
    
    private:
        struct Bucket { int value; unordered_set<string> keys; };
        list<Bucket> buckets;
        unordered_map<string, list<Bucket>::iterator> bucketOfKey;
    };