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].

思路:先用vector记录每个字母出现的位置,然后遍历每个单词,如果每个字母的位置是递增的那么这个单词就符合要求。如何判断位置见的递增关系呢?可以用二分实现。具体看代码

class Solution {
public:
int numMatchingSubseq(string S, vector<string>& words) {
vector<int> v[26];
for (int i = 0; i < S.size(); ++i) {
v[S[i]-'a'].emplace_back(i);
}
int ans = 0;
for (auto i : words) {
int mark = 0;
int y = -1;
for(auto c : i) {
int x = c - 'a';
if (v[x].empty()) {
mark = 1; break;
}
int pos = upper_bound(v[x].begin(), v[x].end(), y) - v[x].begin();
if (pos < v[x].size()) {
y = v[x][pos];
} else {
mark = 1; break;
}
}
if (!mark) ans++;
}
return ans;
}
};

leetcode 792. Number of Matching Subsequences的更多相关文章

  1. 【LeetCode】792. Number of Matching Subsequences 解题报告(Python)

    [LeetCode]792. Number of Matching Subsequences 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://f ...

  2. 792. Number of Matching Subsequences

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

  3. LeetCode 792. 匹配子序列的单词数(Number of Matching Subsequences)

    792. 匹配子序列的单词数 792. Number of Matching Subsequences 相似题目 392. 判断子序列

  4. [LeetCode] Number of Matching Subsequences 匹配的子序列的个数

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

  5. 74th LeetCode Weekly Contest Valid Number of Matching Subsequences

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

  6. [Swift]LeetCode792. 匹配子序列的单词数 | Number of Matching Subsequences

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

  7. LeetCode之“动态规划”:Distinct Subsequences

    题目链接 题目要求: Given a string S and a string T, count the number of distinct subsequences of T in S. A s ...

  8. LeetCode(115) Distinct Subsequences

    题目 Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequen ...

  9. C#版 - Leetcode 191. Number of 1 Bits-题解

    版权声明: 本文为博主Bravo Yeung(知乎UserName同名)的原创文章,欲转载请先私信获博主允许,转载时请附上网址 http://blog.csdn.net/lzuacm. C#版 - L ...

随机推荐

  1. AC日记——Aragorn's Story HDU 3966

    Aragorn's Story Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...

  2. AC日记——[USACO11DEC]牧草种植Grass Planting 洛谷 P3038

    题目描述 Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional road ...

  3. 咦?Oracle归档文件存哪了?

    实验环境:RHEL 5.4 + Oracle 11.2.0.3 现象:日志切换后没找到归档日志目录. 1.查看归档日志路径 2.日志切换后并未找到归档目录 3.创建归档目录后再次观察 引申知识 1.查 ...

  4. Classical method of machine learning

    PCA principal components analysis kmeans bayes spectral clustering svm EM hidden Markov models deep ...

  5. FMDB使用Cached Statement功能

    FMDB使用Cached Statement功能   在FMDB中,Cached Statement功能是一种提高SQLite数据库访问的技术.在SQLite中,所有的SQL语句都会被编译,形成预处理 ...

  6. 偏执的iOS逆向研究员:收集全版本的macOS iOS+越狱+内核调试

    Intro 虽然“只有偏执狂才能够生存”这句话已经被假药停给毁了,但是作为一只有逼格的高大上的iOS逆向分析研究员,难道如果有现成的macOS/iOS全版本镜像可以下载并且无限“漫游”,难道你就不想来 ...

  7. Rebuild account Windows 10 in Domain

    cmd ‘administrator’Regedit 1. Check User Profiles HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows NT\C ...

  8. nodejs 打印机打印 pos打印

    https://www.npmjs.com/package/chn-escpos 安装window vsbuild 编译工具 npm install --global --production win ...

  9. android RecycleView复杂多条目的布局

    用RecycleView来实现布局形式.默认仅仅能指定一种布局格式.可是实际中我们的布局常常会用到多种类型的布局方式.怎样实现呢? 今天来说下经常使用的2钟方式. 第一种: 通过自己定义addHead ...

  10. 介绍一个开源的SIP(VOIP)协议库PJSIP

    本文系转载,出处不可考. 假设你对SIP/VoIP技术感兴趣,哪希望你不要错过:),假设你对写出堪称优美的Code感兴趣 ,那么你也不可错过:) 这期间我想分析一下一个实际的协议栈的设计到实现的相关技 ...