LeetCode 792. Number of Matching Subsequences / 1055. Shortest Way to Form String

792. Number of Matching Subsequences

Inverted Index + Binary Search

Brute Force 会 TLE，时间复杂度为 O(S*n + Σwords[i])。

能够对S中每一个字符创建一个vector，保存其下标位置。这样就能够经过二分快速找到当前位置以后第一次出现的特定字符的下标。时间复杂度 O(S + Σ (words[i]*log(S)) )。

class Solution { public: int numMatchingSubseq(string S, vector<string>& words) { vector<vector<int>> pos(128); // pos[ch] is vector of indexes of ch in S
        for (int i=0;i<S.size();++i) pos[S[i]].push_back(i); int count=0; for (string word:words){ int prev=-1; for (char ch:word){ auto &index_vec=pos[ch]; auto it=upper_bound(index_vec.begin(),index_vec.end(),prev); if (it==index_vec.end()) goto outer; prev = *it; } ++count; outer:; } return count; } };

 

Inverted Index + Table

若是要更快，能够创建一个表 table[i][ch]，记录S的第i个位置后，第一次出现ch的下标。表的构建能够从后往前扫描S，线性时间获得。时间复杂度 O(S + Σwords[i])，牺牲空间换时间。

class Solution { public: int numMatchingSubseq(string S, vector<string>& words) { // after_pos[i][ch] - the index of first ch after ith position of S
        vector<vector<int>> after_pos(S.size(),vector<int>(26,-1)); vector<int> first_pos(26,-1); // first_pos[ch] - the index of first ch in S
        
        for (int i=S.size()-1;i>=0;--i){ char ch=S[i]; for (int j=0;j<26;++j) after_pos[i][j] = first_pos[j]; first_pos[ch-'a'] = i; } int count=0; for (string word:words){ int i=0; int pos=first_pos[word[i++]-'a']; if (pos==-1) goto outer; while (i<word.size()){ pos = after_pos[pos][word[i++]-'a']; if (pos==-1) goto outer; } ++count; outer:; } return count; } };

 

Solution 的方法很像 Trie Tree，对全部word一块儿处理。时间复杂度也是 O(S + Σwords[i])。

https://leetcode.com/problems/number-of-matching-subsequences/solution/

 

1055. Shortest Way to Form String

和上一题如出一辙，一样能够二分或者直接建表作。

class Solution { public: int shortestWay(string source, string target) { vector<vector<int>> pos(26); // pos[ch] - vector of indexes
        for (int i=0;i<source.size();++i) pos[source[i]-'a'].push_back(i); int count=0; int i=0, prev=-1; while (i<target.size()){ char ch=target[i]; auto &index_vec=pos[ch-'a']; if (index_vec.size()==0) return -1; auto it=upper_bound(index_vec.begin(),index_vec.end(),prev); if (it==index_vec.end()){ prev = -1; ++count; }else{ prev = *it; ++i; } } ++count; return count; } };