字符串匹配
1.正则表达式匹配
'.' Matches any single character.
'*' Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).
The function prototype should be:
bool isMatch(const char *s, const char *p)
Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a") → true
isMatch("aa", ".") → true
isMatch("ab", ".") → true
isMatch("aab", "ca*b") → true
这道题用DP可以解决,我们设一个字符串s长度为m,另一个字符串p长度为n
则创建数组dp[m+1][n+1],dp[i][j]表示s字符串[0,i)是否能匹配p字符串[0,j),接下来就是三种情况
class Solution {
public:
bool isMatch(string s, string p) {
int m = s.size(),n=p.size();
vector<vector<bool>>dp(m+1,vector<bool>(n+1,false));
dp[0][0]=true;
for(int i =0;i<=m;++i)
for(int j = 1;j<=n;++j)
{
if(p[j-1]=='*')
dp[i][j]=dp[i][j-2]||(i>0&&(s[i-1]==p[j-2]||p[j-2]=='.')&&dp[i-1][j]); //1.*匹配数量为0 2.s[i-1]与p[j-2]匹配,若匹配,则 dp[i][j]=dp[i-1][j]
else
dp[i][j]=i>0&&(s[i-1]==p[j-1]||p[j-1]=='.')&&dp[i-1][j-1]; //3.若不是'*',则只有一种情况,即s[i-1]==p[j-1]||p[j-1]=='.'
}
return dp[m][n];
}
};
2.通识符匹配
Implement wildcard pattern matching with support for '?' and '*'.
'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).
The matching should cover the entire input string (not partial).
The function prototype should be:
bool isMatch(const char *s, const char *p)
Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "") → true
isMatch("aa", "a") → true
isMatch("ab", "?") → true
isMatch("aab", "ca*b") → false
第一种方法是动态规划,动态方程很好写出
class Solution {
public:
bool isMatch(string s, string p) {
int m = s.size(),n= p.size();
vector<vector<bool>>dp(m+1,vector<bool>(n+1,false));
dp[0][0]=1;
for(int i = 0;i<=m;++i)
for(int j = 1;j<=n;++j)
{
if(p[j-1]=='*')
dp[i][j]=(i>0&&dp[i-1][j])||dp[i][j-1];
else
dp[i][j]=i>0&&(p[j-1]=='?'||p[j-1]==s[i-1])&&dp[i-1][j-1];
}
return dp[m][n];
}
};
第二种方法是贪心
class Solution {
public:
bool isMatch(string s, string p) {
int star=-1,i=0,j=0,ss=0;
while(i<s.size())
{
if(s[i]==p[j]||p[j]=='?')
{
i++;
j++;
}
else if(p[j]=='*')
{
star=j++;
ss=i;
}
else if(star!=-1)
{
j=star+1;
i=++ss;
}
else
return false;
}
while(p[j]=='*')j++;
return j==p.size();
}
};
贪心方法比动态规划快上不少,因为贪心方法可以在遍历的过程中匹配不成功即终止算法
而动态规划需要遍历完,如果数据比较强的话动态规划容易卡住
3.子序列的匹配
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".
第一种方法很容易想出,即贪心的算法,但是被很长且重复的子字符串给卡住时间了,所以用map保存已匹配过的字符串
class Solution {
public:
int numMatchingSubseq(string S, vector<string>& words) {
int res=0;
map<string,int>m;
for(int i = 0;i<words.size();++i)
{
if(m.find(words[i])!=m.end())
{
res+=m[words[i]];
continue;
}
int j,k;
for(j =0,k=0;k<S.size()&&j<words[i].size();++k)
{
if(words[i][j]==S[k])
j++;
}
if(j==words[i].size())
{
res++;
m[words[i]]=1;
}
else
m[words[i]]=0;
}
return res;
}
};
第二种为前缀的想法,匹配了前缀,去掉第一位继续放入queue,随后继续匹配,若长度为1,则结果加1
class Solution {
public:
int numMatchingSubseq(string S, vector<string>& words) {
int res=0;
map<char,queue<string>>m;
for(auto &word:words)
m[word[0]].push(word);
for(auto &c:S)
{
if(m.count(c))
{
int n = m[c].size();
while(n--)
{
string word=m[c].front();
m[c].pop();
if(word.size()==1)
res++;
else
m[word[1]].push(word.substr(1));
}
}
}
return res;
}
};