291. 单词规律 II 🔒

题目描述

给你一种规律 pattern 和一个字符串 s，请你判断 s 是否和 pattern 的规律相匹配。

如果存在单个字符到非空字符串的 双射映射 ，那么字符串 s 匹配 pattern ，即：如果 pattern 中的每个字符都被它映射到的字符串替换，那么最终的字符串则为 s 。双射意味着映射双方一一对应，不会存在两个字符映射到同一个字符串，也不会存在一个字符分别映射到两个不同的字符串。

示例 1：

输入：pattern = "abab", s = "redblueredblue"
输出：true
解释：一种可能的映射如下：
'a' -> "red"
'b' -> "blue"

示例 2：

输入：pattern = "aaaa", s = "asdasdasdasd"
输出：true
解释：一种可能的映射如下：
'a' -> "asd"

示例 3：

输入：pattern = "aabb", s = "xyzabcxzyabc"
输出：false

提示：

1 <= pattern.length, s.length <= 20
pattern 和 s 由小写英文字母组成

解法

方法一

Python3

class Solution:
    def wordPatternMatch(self, pattern: str, s: str) -> bool:
        def dfs(i, j):
            if i == m and j == n:
                return True
            if i == m or j == n or n - j < m - i:
                return False
            for k in range(j, n):
                t = s[j : k + 1]
                if d.get(pattern[i]) == t:
                    if dfs(i + 1, k + 1):
                        return True
                if pattern[i] not in d and t not in vis:
                    d[pattern[i]] = t
                    vis.add(t)
                    if dfs(i + 1, k + 1):
                        return True
                    d.pop(pattern[i])
                    vis.remove(t)
            return False

        m, n = len(pattern), len(s)
        d = {}
        vis = set()
        return dfs(0, 0)

Java

class Solution {
    private Set<String> vis;
    private Map<Character, String> d;
    private String p;
    private String s;
    private int m;
    private int n;

    public boolean wordPatternMatch(String pattern, String s) {
        vis = new HashSet<>();
        d = new HashMap<>();
        this.p = pattern;
        this.s = s;
        m = p.length();
        n = s.length();
        return dfs(0, 0);
    }

    private boolean dfs(int i, int j) {
        if (i == m && j == n) {
            return true;
        }
        if (i == m || j == n || m - i > n - j) {
            return false;
        }
        char c = p.charAt(i);
        for (int k = j + 1; k <= n; ++k) {
            String t = s.substring(j, k);
            if (d.getOrDefault(c, "").equals(t)) {
                if (dfs(i + 1, k)) {
                    return true;
                }
            }
            if (!d.containsKey(c) && !vis.contains(t)) {
                d.put(c, t);
                vis.add(t);
                if (dfs(i + 1, k)) {
                    return true;
                }
                vis.remove(t);
                d.remove(c);
            }
        }
        return false;
    }
}

C++

class Solution {
public:
    bool wordPatternMatch(string pattern, string s) {
        unordered_set<string> vis;
        unordered_map<char, string> d;
        return dfs(0, 0, pattern, s, vis, d);
    }

    bool dfs(int i, int j, string& p, string& s, unordered_set<string>& vis, unordered_map<char, string>& d) {
        int m = p.size(), n = s.size();
        if (i == m && j == n) return true;
        if (i == m || j == n || m - i > n - j) return false;
        char c = p[i];
        for (int k = j + 1; k <= n; ++k) {
            string t = s.substr(j, k - j);
            if (d.count(c) && d[c] == t) {
                if (dfs(i + 1, k, p, s, vis, d)) return true;
            }
            if (!d.count(c) && !vis.count(t)) {
                d[c] = t;
                vis.insert(t);
                if (dfs(i + 1, k, p, s, vis, d)) return true;
                vis.erase(t);
                d.erase(c);
            }
        }
        return false;
    }
};

Go

func wordPatternMatch(pattern string, s string) bool {
	m, n := len(pattern), len(s)
	vis := map[string]bool{}
	d := map[byte]string{}
	var dfs func(i, j int) bool
	dfs = func(i, j int) bool {
		if i == m && j == n {
			return true
		}
		if i == m || j == n || m-i > n-j {
			return false
		}
		c := pattern[i]
		for k := j + 1; k <= n; k++ {
			t := s[j:k]
			if v, ok := d[c]; ok && v == t {
				if dfs(i+1, k) {
					return true
				}
			}
			if _, ok := d[c]; !ok && !vis[t] {
				d[c] = t
				vis[t] = true
				if dfs(i+1, k) {
					return true
				}
				delete(d, c)
				vis[t] = false
			}
		}
		return false
	}
	return dfs(0, 0)
}

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