30. Substring with Concatenation of All Words *HARD*
You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]
You should return the indices: [0,9]
.
(order does not matter).
vector<int> findSubstring(string S, vector<string> &L) { vector<int> result;
if ( S.size()<= || L.size() <= ){
return result;
} int n = S.size(), m = L.size(), l = L[].size(); //put all of words into a map
map<string, int> expected;
for(int i=; i<m; i++){
if (expected.find(L[i])!=expected.end()){
expected[L[i]]++;
}else{
expected[L[i]]=;
}
} for (int i=; i<l; i++){
map<string, int> actual;
int count = ; //total count
int winLeft = i;
for (int j=i; j<=n-l; j+=l){
string word = S.substr(j, l);
//if not found, then restart from j+1;
if (expected.find(word) == expected.end() ) {
actual.clear();
count=;
winLeft = j + l;
continue;
}
count++;
//count the number of "word"
if (actual.find(word) == actual.end() ) {
actual[word] = ;
}else{
actual[word]++;
}
// If there is more appearance of "word" than expected
if (actual[word] > expected[word]){
string tmp;
do {
tmp = S.substr( winLeft, l );
count--;
actual[tmp]--;
winLeft += l;
} while(tmp!=word);
} // if total count equals L's size, find one result
if ( count == m ){
result.push_back(winLeft);
string tmp = S.substr( winLeft, l );
actual[tmp]--;
winLeft += l;
count--;
} }
} return result;
}
用map容器。
i从0到L-1。
actual[word] > expected[word]时舍弃前面的单词,向后查找。
30. Substring with Concatenation of All Words *HARD*的更多相关文章
- LeetCode - 30. Substring with Concatenation of All Words
30. Substring with Concatenation of All Words Problem's Link --------------------------------------- ...
- [Leetcode][Python]30: Substring with Concatenation of All Words
# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 30: Substring with Concatenation of All ...
- [LeetCode] 30. Substring with Concatenation of All Words 解题思路 - Java
You are given a string, s, and a list of words, words, that are all of the same length. Find all sta ...
- leetCode 30.Substring with Concatenation of All Words (words中全部子串相连) 解题思路和方法
Substring with Concatenation of All Words You are given a string, s, and a list of words, words, tha ...
- LeetCode HashTable 30 Substring with Concatenation of All Words
You are given a string, s, and a list of words, words, that are all of the same length. Find all sta ...
- [LeetCode] 30. Substring with Concatenation of All Words 串联所有单词的子串
You are given a string, s, and a list of words, words, that are all of the same length. Find all sta ...
- Java [leetcode 30]Substring with Concatenation of All Words
题目描述: You are given a string, s, and a list of words, words, that are all of the same length. Find a ...
- 【LeetCode】30. Substring with Concatenation of All Words
You are given a string, s, and a list of words, words, that are all of the same length. Find all sta ...
- 【一天一道LeetCode】#30. Substring with Concatenation of All Words
注:这道题之前跳过了,现在补回来 一天一道LeetCode系列 (一)题目 You are given a string, s, and a list of words, words, that ar ...
- [leetcode]30. Substring with Concatenation of All Words由所有单词连成的子串
You are given a string, s, and a list of words, words, that are all of the same length. Find all sta ...
随机推荐
- MySQL备份与恢复-innobackupex
:上一片myloder搞崩溃,为什么百度的博文都是抄袭一模一样的,哎烦! 这一片文章我们来介绍物理备份工具xtracebackup! 首先是安装可以percona官网下载安装,下载rpm包直接yum安 ...
- MySQL索引类型总结和使用技巧
引用地址:http://www.jb51.net/article/49346.htm 在数据库表中,对字段建立索引可以大大提高查询速度.假如我们创建了一个 mytable表: 复制代码 代码如下: C ...
- Python Web学习笔记之Python多线程和多进程、协程入门
进程和线程究竟是什么?如何使用进程和线程?什么场景下需要使用进程和线程?协程又是什么?协程和线程的关系和区别有哪些? 程序切换-CPU时间的分配 首先,我们的任何一个程序都需要运行在一个操作系统中,如 ...
- c++的class声明及相比java的更合理之处
或许是基于一直以来c/c++头文件声明和cXX实现物理上置于独立文件的考虑,c++中的OO在现实中基本上也是按照声明和实现分离的方式进行管理和编译,如下所示: Base.h #pragma once ...
- c++性能之map实现性能比较
http://www.cnblogs.com/zhjh256/p/6346501.html讲述了基本的map操作,在测试的时候,发现map的性能极为低下,与java相比相差了接近200倍.测试的逻辑如 ...
- [Linux 003]——用户和用户组以及 Linux 权限管理(一)
嗬!没想到吧!学习 Linux 的第三天,我们已经开始接触用户管理,用户组管理,以及权限管理这几个逼格满满的关键字.这几个关键字对于前端程序猿的我来说真的是很高大上有木有,以前尝试学 Linux 的时 ...
- Django组件(四) Django之Auth模块
Auth模块概述 Auth模块是Django自带的用户认证模块: 我们在开发一个网站的时候,无可避免的需要设计实现网站的用户系统.此时我们需要实现包括用户注册.用户登录.用户认证.注销.修改密码等功能 ...
- linux下sz rz的正确用法
一.背景 2018年5月30日,今天遇到一个关于串口协议相关的问题,其中涉及到串口传输工具sz,rz等的使用,从man手册中并没有获取到有效信息,因此经过一番搜索,才知这两个工具应该这样使用 二.使用 ...
- javaweb项目运行时生成的Servers项目作用
在javaweb项目中,看到有一个Servers的项目,发现每新增一个项目,就会在Servers项目中新生成一些对应的项目文件. 如图所示: 每个项目都有对应的文件.文件的结构图如下: 解释一:Ser ...
- Python CSV Reader/Writer 例子--转载
CSV(comma-separated values) 是跨多种形式导入导出数据的标准格式,比如 MySQL.Excel. 它以纯文本存储数和文本.文件的每一行就代表一条数据,每条记录包含了由逗号分隔 ...