Description

Given string S and a dictionary of words words, find the number of words[i] that is a subsequence of S.

Example :
Input:

S = "abcde"
words = ["a", "bb", "acd", "ace"]

Output:

3

Explanation:

There are three words in words that are a subsequence of S: "a", "acd", "ace".

Note:

• All words in words and S will only consists of lowercase letters.
• The length of S will be in the range of [1, 50000].
• The length of words will be in the range of [1, 5000].
• The length of words[i] will be in the range of [1, 50].

分析

题目的意思是：给定字典words，和一个字符串S，问字典中有多少个字符串是S的字串。

• 用一个26位的数组alpha存储S种每个字符的索引。然后对于words，用二分法查找每个word中每个字符c的索引是否存在，如果二分查找找到了尾部，说明索引不存在，则这个单词就不是S的substring。
• upper_bound(a.begin(),a.end(),x)返回的是迭代器

代码

class Solution {
public:
    int numMatchingSubseq(string S, vector<string>& words) {
        vector<vector<int>> alpha(26);
        for(int i=0;i<S.size();i++){
            alpha[S[i]-'a'].push_back(i);
        }
        int res=0;
        for(auto word:words){
            int x=-1;
            bool found=true;
            for(char c:word){
                auto it=upper_bound(alpha[c-'a'].begin(),alpha[c-'a'].end(),x);
                if(it==alpha[c-'a'].end()){
                    found=false;
                    break;
                }else{
                    x=*it;
                }
            }
            if(found){
                res++;
            }
        }
        return res;
    }
};

参考文献

​​792. Number of Matching Subsequences​​​​C++STL中的upper_bound()函数的使用​​