LeetCode | Wildcard Matching

最新推荐文章于 2025-05-01 20:06:47 发布

Nero_pf

最新推荐文章于 2025-05-01 20:06:47 发布

阅读量4k

点赞数 1

分类专栏： LeetCode 文章标签： LeetCode java 贪心

本文链接：https://blog.youkuaiyun.com/perfect8886/article/details/22689147

版权

LeetCode 专栏收录该内容

142 篇文章

订阅专栏

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

分析

通过率很低的一题，Java没指针写起来是不顺畅。

贪心的策略，能匹配就一直往后遍历，匹配不上了就看看前面有没有'*'来救救场，再从'*'后面接着试。

代码

public class WildcardMatching {
	public boolean isMatch(String s, String p) {
		int i = 0;
		int j = 0;
		int star = -1;
		int mark = -1;
		while (i < s.length()) {
			if (j < p.length()
					&& (p.charAt(j) == '?' || p.charAt(j) == s.charAt(i))) {
				++i;
				++j;
			} else if (j < p.length() && p.charAt(j) == '*') {
				star = j++;
				mark = i;
			} else if (star != -1) {
				j = star + 1;
				i = ++mark;
			} else {
				return false;
			}
		}
		while (j < p.length() && p.charAt(j) == '*') {
			++j;
		}
		return j == p.length();
	}
}