44. Wildcard Matching -- LeetCode

题目：

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

思路1 动态规划

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 = 1; i <= m; i++)            dp[i][0] = false;        for(int j = 1; j <= n; j++)            dp[0][j] = dp[0][j - 1] && p[j - 1] == '*';        for(int i = 1; i <= m; i++)            for(int j = 1; j <= n; j++)            {                if(p[j - 1] == '*')                    dp[i][j] = dp[i - 1][j - 1] || dp[i - 1][j] || dp[i][j - 1];                else                   dp[i][j] = dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '?') ;            }        return dp[m][n];    }};

思路2 回溯算法

当s[i] == p[j]时i++ j++

当p[j]=='*'时，用iStar和jStar记录当前的i，j

当p[j] != s[i] && p[j] != '?'时，回溯，j回溯到jstar+1，istar++；i=istar

最后就是简单的判断了

小节

思路2的算法思想较简单，但是用代码实现就要小心好多坑了，花了不少时间（看来目前写代码功力并不6啊）

Tips：对于不好理解的算法，可以用编辑器在每次循环将关键的变量打印出来（cout<<'i='<<i ... ...）,帮助直观理解。（ 抽象-》直观）

bool isMatch(string s, string p) {        int  slen = s.size(), plen = p.size(), i, j, iStar=-1, jStar=-1;        for(i=0,j=0 ; i<slen; ++i, ++j)        {            if(p[j]=='*')            { //meet a new '*', update traceback i/j info                iStar = i;                jStar = j;                --i;            }            else            {                 if(p[j]!=s[i] && p[j]!='?')                {  // mismatch happens                    if(iStar >=0)                    { // met a '*' before, then do traceback                        i = iStar++;                        j = jStar;                    }                    else return false; // otherwise fail                }            }        }        while(p[j]=='*') ++j;        return j==plen;    }