【字符串匹配】KMP（implement strStr()), 正则匹配(Wildcard Matching)，2-dim 动规（regular expression)

1 KMP

 

Implement strStr()

 Total Accepted: 15450 Total Submissions: 71217My Submissions

Implement strStr().

Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack.

class Solution {public:    char *strStr(char *haystack, char *needle) {        int n=strlen(needle);        if(n==0)  return haystack;         //kmp        vector<int> nxt(n+1);///@error: nxt should be sized n+1, not n. to contain the nxt[n]        nxt[0]=-1;nxt[1]=0;        for(int i=0;i<n;i++){            int j=nxt[i];            while(j>=0&&needle[i]!=needle[j]) j=nxt[j]; //i matched j            nxt[i+1]=j+1;/////@@@@@@@error: nxt[i+1]=j+1; mistake as needle[i+1]=j+1;  an Array_Variable_Fault !!!!        }        int j=0;        for(int i=0;haystack[i]!=0;i++,j++){            while(j>=0&&haystack[i]!=needle[j]) j=nxt[j];//i matched j            if(j==n-1) return haystack+i-j; //matche needle        }        return NULL;    }};

 正则匹配不同于一般匹配。回溯取决于*号出现的位置。

 采用贪心法：p中'*'尽量匹配s中最少的字符，从上到下执行规则：

    1.1 如果p到达0,s未到达0: 则回溯或匹配失败

    1.2 如果p到达0,s到达0: 则成功

    2.1 如果p为'*':设置回溯点，p前进

    3.1 如果p不为0或'*',而s到达0: 则匹配失败

    3.2 如果p不为0或'*',s不为0: p=s或p='?'则s、p前进；否则回溯或匹配失败。

Wildcard Matching

 Total Accepted: 10189 Total Submissions: 75093My Submissions

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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false

class Solution {public:    bool isMatch(const char *s, const char *p) {       const char *pre_s,*pre_p;bool star=false;///@@error:pre_s pre_p must be const, or invalid conversion from 'const char*' to 'char*' [-fpermissive]       while(true){           if(*p==0){             if(*s==0) return true;  //*p=0 and *p matches *s             if(!star) return false; //*p=0!=*s             s=++pre_s;p=pre_p;           }           else if(*p!='*'){                if(*p==*s||*p=='?'&&*s!=0) s++,p++;//*p!=0 and *p matches *s (then *s!=0)                else {                    if(!star||*s==0) return false;///@warning:when *s=0,say s reaches the end,p will not match it even if Star=true                    s=++pre_s;p=pre_p;                }           }else{                star=true;                pre_s=s,pre_p=p+1;                p=pre_p;           }       }    }};

或者

class Solution {public: bool isMatch(const char *s, const char *p) {    // Start typing your C/C++ solution below    // DO NOT write int main() function    const char *pre_s = s, *pre_p = p;//保存上一次开始匹配的位置    bool has_star = false;    while (*s) {      if (*s != *p) {        if (*p == '?') s++, p++;        else if (*p == '*') {          has_star = true;          while (*p == '*') p++;          if (*p == '\0') return true;          pre_s = s, pre_p = p;//置上一个开始比较的位置        } else if (has_star) {          pre_s++;          s = pre_s, p = pre_p;//恢复到上一次比较的下一个位置        } else return false;   //*s!=0 and *s!=*p and *p!='?' and !hasStar      } else s++, p++;    }    //s has reach the end, so p's remainning part must match ''.    while (*p == '*') p++;    return *p == '\0';  }};

或者：

class Solution {public:    bool isMatch(const char *s, const char *p) {        bool tag = false;         const char * pre_s, *pre_p;        while(*s){            if(*s==*p || *p=='?') s++, p++;            else if(*p=='*'){                tag = true;                pre_s = s, pre_p = ++p;            }            else{                if(tag){                s = ++ls, p = pre_p;                }                else return false;            }        }        while(*p == '*') p++;        return *p==0;    }};

Regular Expression Matching

 Total Accepted: 13590 Total Submissions: 68280My Submissions

Implement regular expression matching with support for '.' and '*'.

'.' 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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

一个很典型的二维动规的实例
for(char in Pattern)
     for(char in String)

class Solution {public:    inline bool match(char a,char b){        //assert(a!='*');        if(a=='.') return true;        return a==b;    }    bool isMatch(const char *s, const char *p) {        int n=strlen(p),m=strlen(s);        vector<vector<bool> > a(n+1,vector<bool>(m+1,false));        a[0][0]=true;        /*        for(int i=2;i<=n;i+=2){            if(p[i-1]!='*') break;            a[i][0]=true;        }        for(int j=1;j<=m;j++)            a[0][j]=false;        */        for(int i=1;i<=n;i++){            if(i<n&&p[i]=='*') continue;//i++,go to '*'            if(p[i-1]=='*'){                assert(i>1&&p[i-2]!='*');                char c=p[i-2];                bool lc=false;                for(int j=0;j<=m;j++){//1-dim dp                    if(j>0&&lc) lc=match(c,s[j-1]);///@error:s[j-1] fault as a[j-1]///@error:if(j>0&&lc&&match(c,s[j-1])) lc=true; here,when lc=true, lc should be false if match(c,s[j-1])=false                    if(a[i-2][j]) lc=true;                    a[i][j]=lc;                }                        }            else               for(int j=1;j<=m;j++){                    if((p[i-1]==s[j-1]||p[i-1]=='.')&& a[i-1][j-1])  a[i][j]=true;             }        }        return a[n][m];    }};