刷题10. Regular Expression Matching
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一、题目说明
这个题目是10. Regular Expression Matching,乍一看不是很难。
但我实现提交后,总是报错。不得已查看了答案。
二、我的做法
我的实现,最大的问题在于对.*
的处理有问题,始终无法成功。
#include<iostream>
using namespace std;
class Solution{
public:
bool isMatch(string s,string p){
bool result = true;
if(s.length()<=0 && p.length()<=0){
return true;
}
if(p==".*"){
return true;
}
int sCurr=0,pCurr=0;
int lenS = s.length();
int lenP = p.length();
//count the num of .*
int numOfWildCard = 0;
while(pCurr<lenP){
if(p[pCurr]=='.' && pCurr+1<lenP && p[pCurr+1]=='*'){
numOfWildCard++;
}
pCurr++;
}
//cout<<numOfWildCard<<":";
pCurr = 0;
while(sCurr<lenS && pCurr<lenP){
if((pCurr+1<lenP) && p[pCurr]=='.' && p[pCurr+1]=='*'){
if(pCurr+2<lenP){
pCurr = pCurr+2;
while(sCurr<lenS && s[sCurr]!=p[pCurr]){
sCurr++;
}
}
}
if((pCurr+1<lenP) && p[pCurr+1]=='*'){
while(sCurr<lenS && s[sCurr]==p[pCurr]){
sCurr++;
}
pCurr = pCurr+2;
}
if(sCurr<lenS && pCurr<lenP && p[pCurr+1]!='*'){
if(s[sCurr]==p[pCurr] || p[pCurr]=='.'){
sCurr++;
pCurr++;
}
}
}
if(sCurr==lenS && pCurr==lenP){
result = true;
}else{
result = false;
}
return result;
}
};
int main(){
Solution s;
cout<<(false==s.isMatch("aa","a"))<<endl;
cout<<(true==s.isMatch("aa","a*"))<<endl;
cout<<(true==s.isMatch("ab",".*"))<<endl;
cout<<(true==s.isMatch("aab","c*a*b"))<<endl;
cout<<(false==s.isMatch("mississippi","mis*is*p*."))<<endl;
return 0;
}
三、正确的做法
1.递归方法
#include<iostream>
#include<vector>
using namespace std;
class Solution{
public:
bool isMatch(string s, string p) {//aa a
if(p.empty()) return s.empty();
if(s.empty()) return p.empty() || (p[1] == '*' ? isMatch(s, p.substr(2)) : false);
if(p[0] != '.' && s[0] != p[0]) return p[1] == '*' ? isMatch(s, p.substr(2)) : false;
if(p[1] == '*') return isMatch(s.substr(1), p) || isMatch(s, p.substr(2));
return isMatch(s.substr(1), p.substr(1));
}
};
int main(){
Solution s;
cout<<(false==s.isMatch("aa","a"))<<endl;
cout<<(true==s.isMatch("aa","a*"))<<endl;
cout<<(true==s.isMatch("ab",".*"))<<endl;
cout<<(true==s.isMatch("aab","c*a*b"))<<endl;
cout<<(false==s.isMatch("mississippi","mis*is*p*."))<<endl;
cout<<(false==s.isMatch("ab",".*c"))<<endl;
cout<<(true==s.isMatch("aaa","a*a"))<<endl;
return 0;
}
2.DP方法
dp是什么?动态规划啊,
#include<iostream>
#include<vector>
#include <mem.h>
using namespace std;
class Solution {
public:
bool isMatch(string s, string p) {
int ssize = s.size(),psize = p.size();
string pp="";
vector<bool> star;
for(int i=0;i<p.size();i++){
if(p[i]=='*'){
star.back()=true;
}else{
star.push_back(false);
pp+= p[i];
}
}
psize = pp.size();
bool dp[psize+1][ssize+1];
memset(dp,false,sizeof(dp));
dp[0][0] = true;
for(int i=1;i<=psize;i++){
if(star[i-1]==true){
dp[i][0] = true;
}else{
break;
}
}
for(int i=1;i<=psize;i++)
for(int j=1;j<=ssize;j++){
if(dp[i-1][j-1]== true){
if(pp[i-1]==s[j-1] || pp[i-1]=='.'){
dp[i][j] = true;
continue;
}
}
if(dp[i-1][j]== true){
if(star[i-1]==true){
dp[i][j] = true;
continue;
}
}
if(dp[i][j-1]== true){
if(star[i-1]==true && (pp[i-1]==s[j-1] || pp[i-1]=='.')){
dp[i][j] = true;
continue;
}
}
}
return dp[psize][ssize];
}
};
int main(){
Solution s;
cout<<(false==s.isMatch("aa","a"))<<endl;
cout<<(true==s.isMatch("aa","a*"))<<endl;
cout<<(true==s.isMatch("ab",".*"))<<endl;
cout<<(true==s.isMatch("aab","c*a*b"))<<endl;
cout<<(false==s.isMatch("mississippi","mis*is*p*."))<<endl;
cout<<(false==s.isMatch("ab",".*c"))<<endl;
cout<<(true==s.isMatch("aaa","a*a"))<<endl;
cout<<(false==s.isMatch("a",""))<<endl;
return 0;
}
四、总结
看来基础知识还需要恶补,加油!
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