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‘?‘ 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
思路：使用动态动态规划的解法
设dp[i][j]表示s[0,i]与p[0,j]是否match;
代码如下：这种方法在运行时内存溢出了
public boolean isMatch(String s, String p){ if(s != null && p != null && s.equals(p)) return true; if("".equals(s) || "".equals(p)) return false; int n = s.length(); int m = p.length(); boolean[][] dp = new boolean[n][m]; if(s.charAt(0) == p.charAt(0) || p.charAt(0) == ‘?‘ || p.charAt(0) == ‘*‘) dp[0][0] = true; else return false; for(int i=0; i<n; i++) { for(int j=0; j<m; j++) { if(i == 0 && j > 0) { if(p.charAt(j) == ‘*‘) dp[i][j] = dp[i][j-1]; else dp[i][j] = false; }else if(i > 0 && j == 0) { if(p.charAt(j) == ‘*‘) dp[i][j] = true; else dp[i][j] = false; } else if(i>0 && j>0){ if(p.charAt(j) == ‘*‘) { dp[i][j] = (dp[i-1][j] || dp[i][j-1]); } else if(p.charAt(j) == ‘?‘ || s.charAt(i) == p.charAt(j)) { dp[i][j] = dp[i-1][j-1]; } else { return false; } } } } return dp[n-1][m-1]; }
使用循环遍历两个字符串，代码如下
public boolean isMatch(String s, String p) { if("*".equals(p)) return true; int posS = 0; int posP = 0; int posG = -1;//最近一个“*”的位置 int posSnext = -1; while(posS < s.length()) { if(posP<p.length() && p.charAt(posP) == ‘*‘) { posG = posP; posP++; posSnext = posS + 1; } else if(posP<p.length() && (s.charAt(posS) == p.charAt(posP) || p.charAt(posP) == ‘?‘)) { posS++; posP++; } else if(posG == -1){ return false; } else { posP = posG+1; posS = posSnext; posSnext++; } } while(posP < p.length() && p.charAt(posP) == ‘*‘) posP++; if(posP >= p.length()) return true; return false; }
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原文地址：http://www.cnblogs.com/linxiong/p/4347718.html
