[LeetCode] 291. Word Pattern II 词语模式 II

Given a pattern and a string str, find if str follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty substring in str.

Examples:

1. pattern = "abab", str = "redblueredblue" should return true.
2. pattern = "aaaa", str = "asdasdasdasd" should return true.
3. pattern = "aabb", str = "xyzabcxzyabc" should return false.

Notes:
You may assume both pattern and str contains only lowercase letters.

290. Word Pattern 的拓展，区别是这里的单词字符串没有空格了，不能按空格把单词直接拆分出来。

可以用回溯法来判断每一种情况，用哈希表建立模式字符和单词之间的映射，还需要用变量p和r来记录当前递归到的模式字符和单词串的位置，在递归函数中，如果p和r分别等于模式字符串和单词字符串的长度，说明此时匹配成功结束了，返回ture，反之如果一个达到了而另一个没有，说明匹配失败了，返回false。

解法：回溯Backtracking

Python:

# Time:  O(n * C(n - 1, c - 1)), n is length of str, c is unique count of pattern,
#                                there are H(n - c, c - 1) = C(n - 1, c - 1) possible splits of string,
#                                and each one costs O(n) to check if it matches the word pattern.
# Space: O(n + c)

class Solution(object):
    def wordPatternMatch(self, pattern, str):
        """
        :type pattern: str
        :type str: str
        :rtype: bool
        """
        w2p, p2w = {}, {}
        return self.match(pattern, str, 0, 0, w2p, p2w)

    def match(self, pattern, str, i, j, w2p, p2w):
        is_match = False
        if i == len(pattern) and j == len(str):
            is_match = True
        elif i < len(pattern) and j < len(str):
            p = pattern[i]
            if p in p2w:
                w = p2w[p]
                if w == str[j:j+len(w)]:  # Match pattern.
                    is_match = self.match(pattern, str, i + 1, j + len(w), w2p, p2w)
                # Else return false.
            else:
                for k in xrange(j, len(str)):  # Try any possible word
                    w = str[j:k+1]
                    if w not in w2p:
                        # Build mapping. Space: O(n + c)
                        w2p[w], p2w[p] = p, w;
                        is_match = self.match(pattern, str, i + 1, k + 1, w2p, p2w)
                        w2p.pop(w), p2w.pop(p);
                    if is_match:
                        break
        return is_match　　

C++：

class Solution {
public:
    bool wordPatternMatch(string pattern, string str) {
        unordered_map<char, string> m;
        return helper(pattern, 0, str, 0, m);
    }
    bool helper(string pattern, int p, string str, int r, unordered_map<char, string> &m) {
        if (p == pattern.size() && r == str.size()) return true;
        if (p == pattern.size() || r == str.size()) return false;
        char c = pattern[p];
        for (int i = r; i < str.size(); ++i) {
            string t = str.substr(r, i - r + 1);
            if (m.count(c) && m[c] == t) {
                if (helper(pattern, p + 1, str, i + 1, m)) return true;
            } else if (!m.count(c)) {
                bool b = false;
                for (auto it : m) {
                    if (it.second == t) b = true;
                }
                if (!b) {
                    m[c] = t;
                    if (helper(pattern, p + 1, str, i + 1, m)) return true;
                    m.erase(c);
                }
            }
        }
        return false;
    }
};　　

C++:

class Solution {
public:
    bool wordPatternMatch(string pattern, string str) {
        unordered_map<char, string> m;
        set<string> s;
        return helper(pattern, 0, str, 0, m, s);
    }
    bool helper(string pattern, int p, string str, int r, unordered_map<char, string> &m, set<string> &s) {
        if (p == pattern.size() && r == str.size()) return true;
        if (p == pattern.size() || r == str.size()) return false;
        char c = pattern[p];
        for (int i = r; i < str.size(); ++i) {
            string t = str.substr(r, i - r + 1);
            if (m.count(c) && m[c] == t) {
                if (helper(pattern, p + 1, str, i + 1, m, s)) return true;
            } else if (!m.count(c)) {
                if (s.count(t)) continue;
                m[c] = t;
                s.insert(t);
                if (helper(pattern, p + 1, str, i + 1, m, s)) return true;
                m.erase(c);
                s.erase(t);
            }
        }
        return false;
    }
};

原文地址：https://www.cnblogs.com/lightwindy/p/9736251.html