LeetCode-44 Wildcard Matching
'?' 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]; }
先对匹配串做预处理,使**变成*,使用两个指针分别指向两个字符串,循环遍历。
public boolean isMatch(String s, String p) { p = prepare(p); int pi=0; int lastStar=-1; int lastMatch = 0; for(int si=0; si<s.length(); ) { if(pi<p.length() && (p.charAt(pi) == '?' || s.charAt(si) == p.charAt(pi))) { pi++; si++; } else if(pi<p.length() && p.charAt(pi) == '*') { lastStar = pi; pi++; lastMatch = si; } else if(lastStar != -1){ pi = lastStar+1; lastMatch++; si = lastMatch; } else { return false; } } if(pi == p.length() || pi == p.length()-1 && p.charAt(pi) == '*') return true; return false; } //匹配串预处理 public String prepare(String str) { StringBuilder sb = new StringBuilder(str); for(int i=1; i<sb.length(); ) { if(sb.charAt(i) == '*' && sb.charAt(i) == sb.charAt(i-1)) sb.deleteCharAt(i); else i++; } return sb.toString(); }
posted on 2015-03-18 16:56 linxiong1991 阅读(151) 评论(0) 编辑 收藏 举报