标签：style blog class code c java
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
首先想到的是递归，但是超时，然后递归转非递归，在大数据测试用例时还是超时，这里贴一下非递归的代码
1 public boolean isMatch(String s, String p) { 2 Stack<String> source = new Stack<String>(); 3 Stack<String> regex = new Stack<String>(); 4 if (s.equals("")) { 5 if (p.equals("") || p.equals("*")) { 6 return true; 7 } else { 8 return false; 9 } 10 } 11 char temp[] = p.toCharArray(); 12 p = ""; 13 for (char a : temp) { 14 if (p.length() > 0 && p.charAt(p.length() - 1) == ‘*‘ && a == ‘*‘) { 15 continue; 16 } 17 p += a; 18 } 19 boolean result = false; 20 source.push(s); 21 regex.push(p); 22 while (!result && !source.empty()) { 23 s = source.pop(); 24 p = regex.pop(); 25 int i = 0; 26 int j = 0; 27 for (; i < s.length() && j < p.length();) { 28 if (p.charAt(j) != ‘*‘) { 29 if (p.charAt(0) == ‘?‘ || p.charAt(0) == s.charAt(0)) { 30 i++; 31 j++; 32 } else { 33 break; 34 } 35 } else { 36 j++; 37 if (j == p.length()) { 38 i = s.length(); 39 } else { 40 for (int k = i; k < s.length(); k++) { 41 if (s.charAt(k) == p.charAt(j)) { 42 source.push(s.substring(k)); 43 regex.push(p.substring(j)); 44 } 45 } 46 continue; 47 } 48 } 49 } 50 if (i == s.length() && j == p.length()) { 51 result = true; 52 } 53 } 54 return result; 55 }
优化后还是超时，就想着应该有别的思路。
借鉴一下网上的，转载自 http://blog.csdn.net/pickless/article/details/9787227
看着有点像求字符串的最长公共子串那个问题的解法
用矩阵表示匹配结果，横坐标为p串，纵坐标为s串，假设S="ababa"，P="a*?"，则有下图：
| S\P | a | * | ? |
| a | T | T | F |
| b | F | T | T |
| a | F | T | T |
| b | F | T | T |
如何计算这个表格呢？规则如下：
1. p[j] == ‘*‘，则 f[i][j] = f[i-1][j]
2. p[j] == ‘?‘ || p[j] == s[i] 则 f[i][j] = f[i-1][j-1]
3. 其余情况 f[i][j] = false
上述情况不包含，第0行和第0列，需要单独计算
经过上面分析，第i行只与第i-1行有关，所以，可以将矩阵简化成一维的。代码如下：
1 public boolean isMatch(String s, String p) { 2 boolean f[] = new boolean[p.length()]; 3 if (s == null || p == null || (p.equals("") && !s.equals(""))) { 4 return false; 5 } 6 if (s.equals("")) { 7 if (p.equals("") || p.equals("*")) { 8 return true; 9 } else { 10 return false; 11 } 12 } 13 boolean isTrue = false; 14 for (int i = 0; i < p.length(); i++) { 15 if (p.charAt(i) == ‘*‘) { 16 f[i] = (i == 0) || f[i - 1]; 17 } else if (p.charAt(i) == ‘?‘ || p.charAt(i) == s.charAt(0)) { 18 if (isTrue) { 19 break; 20 } 21 f[i] = (i == 0) || ((!isTrue) && f[i - 1]); 22 isTrue = true; 23 } else { 24 break; 25 } 26 } 27 for (int i = 1; i < s.length(); i++) { 28 boolean temp[] = new boolean[f.length]; 29 for (int j = 0; j < f.length; j++) { 30 if (p.charAt(j) == ‘*‘) { 31 temp[j] = f[j]; 32 } else if (p.charAt(j) == ‘?‘ || p.charAt(j) == s.charAt(i)) { 33 temp[j] = (j == 0 && p.charAt(0) == ‘*‘ && f[j]) 34 || (j > 0 && f[j - 1]); 35 } else { 36 temp[j] = false; 37 } 38 } 39 f = temp; 40 } 41 return f[f.length - 1]; 42 }
但是依旧超时，再根据网上的解法，使用贪心算法
1 public boolean isMatch(String s, String p) { 2 int ls = s.length(); 3 int lp = p.length(); 4 if(ls == 0 && lp == 0) return true; 5 else if(ls != 0 && lp == 0) return false; 6 7 String[] strarr = p.split("[\\*]+",0); 8 if(strarr.length == 0) return true; 9 boolean firstneedmatch = !p.startsWith("*"); 10 boolean lastneedmatch = !p.endsWith("*"); 11 if(!firstneedmatch) strarr = Arrays.copyOfRange(strarr, 1, strarr.length); 12 if(strarr.length == 0) return true; 13 14 int base = 0; 15 for(int i=0;i<strarr.length;i++){ 16 if(base > ls-strarr[i].length()) return false; 17 if(i==0 && firstneedmatch) base = -1; 18 else if(i==strarr.length-1 && lastneedmatch) base = -2; 19 base = getMatchedIdx(s,strarr[i],base); 20 if(base == -1) return false; 21 if(i == strarr.length-1 && lastneedmatch && base != ls) return false; 22 } 23 return true; 24 } 25 26 public int getMatchedIdx(String s, String pat,int base){ 27 char[] sch = s.toCharArray(); 28 char[] pch = pat.toCharArray(); 29 int ls = sch.length; 30 int lp = pch.length; 31 if(base == -1){ 32 for(int i=0;i<lp;i++){ 33 if(pch[i] != ‘?‘ && pch[i] != sch[i]) return -1; 34 } 35 return lp; 36 } 37 else if(base == -2){ 38 base = ls-lp; 39 for(int i=0;i<lp;i++){ 40 if(pch[i] !=‘?‘ && pch[i] != sch[base+i]) return -1; 41 } 42 return ls; 43 } 44 else{ 45 int limit = ls-lp; 46 for(;base<=limit;base++){ 47 for(int i=0;i<lp;i++){ 48 if(pch[i] !=‘?‘ && pch[i] != sch[base+i]) break; 49 if(i == lp-1) return base+lp; 50 } 51 } 52 return -1; 53 } 54 }
Wildcard Matching,布布扣,bubuko.com
标签：style blog class code c java
原文地址：http://www.cnblogs.com/apoptoxin/p/3731530.html
