标签：leetcode substring map 窗口移动法 字符串匹配
【题目】
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 wordsexactly 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).
【分析】
本题目解法有两个，第一个是常规思路，第二个是窗口移动法，这个方法需要掌握。
解法一：先把words存在一个map中，key是单词字符串，value是出现的次数。然后逐个位置遍历字符串s（注意遍历结束位置不必到最后，剩余长度小于单词总长度即停止），判断其后面的和档次总长度相同的子串中的每个单词是否和words一样。如果一样，这push_back进去；否则，遍历下一个字符。时间复杂度：O(LNw)，L是s的长度，N是words的个数，w是word的长度。
代码：
class Solution {
public:
vector<int> findSubstring(string s, vector<string>& words) {
vector <int> res;
int N = words.size(); // number of words
if(N == 0)
return res;
int len = words[0].length(); // length of each word
int strLen = s.length(); // length of string
map<string, int> countWords; // get the wordCount map
for(int i = 0; i < N; i ++)
{
if(countWords.count(words[i]))
countWords[words[i]]++;
else
countWords[words[i]] = 1;
}
map<string, int> counting;
for(int i = 0; i <= strLen - len * N; i++) // first loop
{
counting.clear();
bool flag = true;
for(int j = i; j < i + N*len; j += len) // second loop
{
string w = s.substr(j, len);
if(countWords.count(w) == 0) // if not exist in countWords, break directly
{
flag = false;
break;
}
else
{
if(counting.count(w)) // if not new
counting[w]++;
else
counting[w] = 1;
}
}
if(!flag)
continue;
else
{
if(compare(counting, countWords))
res.push_back(i);
else
continue;
}
}
}
bool compare(map<string, int> counting, map<string, int> countWords)
{
map<string, int>::iterator iter;
for(iter = countWords.begin(); iter != countWords.end(); iter++)
{
if(counting[iter->first] != iter->second)
return false;
}
return true;
}
};解法二：窗口移动法。
首先，初始化一个长度为0的窗口，定义头部是begin，尾部是tail。判断tail后面的一个单词，如果在words里面，则tail往后移动，即窗口伸长一个单词的量；如果tail后面的单词压根儿不在words里面，那么把begin后移到最后位置，重新初始化窗口，继续判断；如果tail后面的单词是之前出现过，不过words中没有这个单词的容量，那么begin后移到该单词第一次出现的位置，继续判断。这个方法的时间复杂度为O(Lw)。
代码：
class Solution {
public:
vector<int> findSubstring(string S, vector<string> &L) {
unordered_map<string, int>wordTimes;//L中单词出现的次数
for(int i = 0; i < L.size(); i++)
if(wordTimes.count(L[i]) == 0)
wordTimes.insert(make_pair(L[i], 1));
else wordTimes[L[i]]++;
int wordLen = L[0].size();
vector<int> res;
for(int i = 0; i < wordLen; i++)
{//为了不遗漏从s的每一个位置开始的子串，第一层循环为单词的长度
unordered_map<string, int>wordTimes2;//当前窗口中单词出现的次数
int winStart = i, cnt = 0;//winStart为窗口起始位置,cnt为当前窗口中的单词数目
for(int winEnd = i; winEnd <= (int)S.size()-wordLen; winEnd+=wordLen)
{//窗口为[winStart,winEnd)
string word = S.substr(winEnd, wordLen);
if(wordTimes.find(word) != wordTimes.end())
{
if(wordTimes2.find(word) == wordTimes2.end())
wordTimes2[word] = 1;
else wordTimes2[word]++;
if(wordTimes2[word] <= wordTimes[word])
cnt++;
else
{//当前的单词在L中，但是它已经在窗口中出现了相应的次数，不应该加入窗口
//此时，应该把窗口起始位置想左移动到，该单词第一次出现的位置的下一个单词位置
for(int k = winStart; ; k += wordLen)
{
string tmpstr = S.substr(k, wordLen);
wordTimes2[tmpstr]--;
if(tmpstr == word)
{
winStart = k + wordLen;
break;
}
cnt--;
}
}
if(cnt == L.size())
res.push_back(winStart);
}
else
{//发现不在L中的单词
winStart = winEnd + wordLen;
wordTimes2.clear();
cnt = 0;
}
}
}
return res;
}
};class Solution {
public:
vector<int> findSubstring(string s, vector<string>& words) {
vector <int> res;
int N = words.size(); // number of words
if(N == 0)
return res;
int len = words[0].length(); // length of each word
int strLen = s.length(); // length of string
map<string, int> countWords; // get the wordCount map
for(int i = 0; i < N; i ++)
{
if(countWords.count(words[i]))
countWords[words[i]]++;
else
countWords[words[i]] = 1;
}
map<string, int> counting = countWords;
int begin = 0, tail = begin;
while(tail < strLen)
{
if(tail - begin + 1 == N * len) // get results
{
res.push_back(begin);
begin = tail + 1;
tail = begin;
counting = countWords;
continue;
}
string w = s.substr(tail, len);
int kind = moveClass(w, counting); // get the manner how to move the tail
if(kind == 1) // if w not in countWords
{
begin = tail + 2;
tail = begin;
counting = countWords;
continue;
}
if(kind == 2) // if w in countWords now
{
tail += len;
continue;
}
if(kind == 3) // if w in countWords in past
{
string tmp = s.substr(begin, tail - begin + 1);
begin = tmp.find(w) + 1;
tail = begin;
counting = countWords;
continue;
}
}
return res;
}
int moveClass(string w, map<string, int>& counting)
{
if(counting.count(w) == 0)
return 1;
else if(counting[w] >= 1)
{
counting[w]--;
return 2;
}
else
return 3;
}
};
Substring with Concatenation of All Words——解题报告（窗口移动法）
标签：leetcode substring map 窗口移动法 字符串匹配
原文地址：http://blog.csdn.net/puqutogether/article/details/45897333
