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
思路一：
代碼如下：
class Solution {public:    bool isMatch(string s, string p) {        int sIndex=0,pIndex=0,match=0,starIdx=-1;        while(sIndex < s.length())        {            if(pIndex < p.length() && (p[pIndex] == '?' || p[pIndex] == s[sIndex]))            {                pIndex++;                sIndex++;            }            else if(pIndex < p.length() && p[pIndex] == '*')            {                starIdx = pIndex;                pIndex++;                match = sIndex;            }            else if(starIdx!=-1)            {                pIndex = starIdx+1;                match++;                sIndex = match;            }            else                return false;        }                while(pIndex<p.length() && p[pIndex] == '*')            pIndex++;                    return pIndex == p.length();    }};
思路二：使用大殺器 -動態規劃，列出動態方程如下：
if p[j-1] != '*', then dp[i][j] = dp[i-1][j-1] && (s[i-1] == p[j-1] || p[j-1] == '?')
if p[j-1] == '*', then dp[i][j] = dp[i-1][j] || dp[i][j-1]
需要注意的是dp方程組的初始化，dp[0][i](i=1,2,3...)根據p來進行，因為*可以代表空，所以當開頭字串為*時，需要在dp[0][i]中相應位置設為true。 最后dp[0][0]設置為true
代碼如下：
class Solution {public:    bool isMatch(string s, string p) {            int m = s.length(),n=p.length();        bool dp[m+1][n+1];        memset(dp,false,sizeof(bool)*(m+1)*(n+1));        dp[0][0] = true;        for(int i=1;i<=n;i++)        {           if(p[i-1] == '*')                dp[0][i] = true;            else                break;        }                                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] && (s[i-1] == p[j-1] || p[j-1] == '?');                else if(p[j-1] == '*')                    dp[i][j] = dp[i-1][j] || dp[i][j-1];            }        return dp[m][n];    }};