[Leetcode] 44. Wildcard Matching

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”) ? false
isMatch(“aa”,”aa”) ? true
isMatch(“aaa”,”aa”) ? false
isMatch(“aa”, “*”) ? true
isMatch(“aa”, “a*”) ? true
isMatch(“ab”, “?*”) ? true
isMatch(“aab”, “c*a*b”) ? false

字符串匹配，”?”任意一个字符，”*”0 个或者多个任意字符
参考leetcode 10 http://blog.csdn.net/neilooo/article/details/77284682

思路 DP

和leetcode 10 基本一样，情况还要简单些。
建立二维数组记录是否匹配，传递真值。
dp[i][j]含义是字符串s[0:i] p[0:j]是否匹配。

值传递情况：

• s[i] p[j]匹配，dp[i][j] = dp[i-1]][j-1]
• p[j] = “*”，dp[i][j] = dp[i-1][j] || dp[i][j-1]
  传递dp[i][j-1]：*代表的字符数为0
  传递dp[i-1][j]：*代表的字符数为 多个

dp二维数组 i,j=0，代表空串
对应s[i-1],p[j-1]才是字符串中的位置。

    bool isMatch(string s, string p) {        int rows = s.size() + 1;        int cols = p.size() + 1;        vector<vector<bool>> dp(rows, vector<bool>(cols, false));        dp[0][0] = true;        for (int i = 1; i < rows; i++){            dp[i][0] = false;        }        for (int j = 1; j < cols; j++){            if (p[j-1] == '*')                dp[0][j] = dp[0][j-1];        }        for (int i = 1; i < rows; i++){            for (int j = 1; j < cols; j++){                if (p[j-1] == s[i-1] || p[j-1] == '?'){                    dp[i][j] = dp[i-1][j-1];                }                else if (p[j-1] == '*'){                    dp[i][j] = dp[i-1][j] || dp[i][j-1];                }            }        }        return dp[rows-1][cols-1];    }