算法第15周Wildcard Matching[hard]
2017-12-17 19:21
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Description
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
Analysis
这道题目是进行字符串匹配。*符号可以匹配任意的字符串(包括空字符串),?符号可以匹配任意字符。我们从左到右依次扫描字符串和模板。 如果遇到‘*’号,我们需要记录下此时所处的字符串位置istar和模板位置jstar,并将模板移后一位继续进行匹配。一旦遇到不匹配的情况(p[j] != s[i] && p[j] != '?'),我们考虑之前是否遇到'*'号,如果遇到那我们把遇到‘*’号到现在所匹配的符号都认为是与'*'进行匹配成功的,这是我们需要将j移到遇到‘*’号时位置,继续进行匹配。 最后还需检查是否扫描完模板。
Solution
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pre class="prettyprint">
class Solution { public: bool isMatch(string s, string p) { int m = s.length(); int n = p.length(); int i, j; int istar, jstar; istar = -1; jstar = -1; for (i = 0, j = 0; i < m;) { if (p[j] == '*') { istar = i; jstar = j; j++; } else { if (p[j] != s[i] && p[j] != '?') { if (istar >= 0) { i = istar++; j = jstar; i++; } else { return false; } } else { i++; j++; } } } while (p[j] == '*') j++; if (j == n) return true; return false; } };
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