[Google] LeetCode 1554 Strings Differ by One Character 哈希

Given a list of strings dict where all the strings are of the same length.

Return true if there are 2 strings that only differ by 1 character in the same index, otherwise return false.

Solution

对于每个字符串，我们用哈希将其映射为数。然后对于每个位置（即删除的位置），我们枚举每个字符串，得到该字符串删去该位置之后的哈希值。这里我们用一个 \(map\) 来存储此时哈希值对应的字符串下标。所以我们就可以在 \(map\) 里面查找其对应的下标，如果相同的话，那么就是 \(true\)

点击查看代码

class Solution {
private:
    int mod = 1e9+7;
    vector<long long> hash;
    int p = 37;
public:
    bool differByOne(vector<string>& dict) {
        int n=dict.size();
        int m = dict[0].size();
        hash = vector<long long> (n,0);
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                hash[i]=(long long)(p*hash[i]+(dict[i][j]-'a'))%mod;
            }
        }
        
        for(long long j=m-1, bs = 1;j>=0;j--){
            // hash_val -> pos
            unordered_map<long long,vector<long long>> mp;
            for(long long i=0;i<n;i++){
                long long res = (mod+hash[i]-bs*(dict[i][j]-'a')%mod)%mod;
                if(mp.find(res)!=mp.end()){
                    auto f = mp.find(res);
                    for(auto ele:f->second){
                        if(equal(begin(dict[i]),begin(dict[i])+j, begin(dict[ele])) && equal(begin(dict[i])+j+1, end(dict[i]), begin(dict[ele])+j+1))
                            return true;
                    }
                }
                mp[res].push_back(i);
            }
            bs=bs*p%mod;
        }
        return false;
    }
};