[leetcode] 290. Word Pattern

Question:

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 word in str.

Examples:pattern = "abba", str = "dog cat cat dog" should return true.pattern = "abba", str = "dog cat cat fish" should return false.pattern = "aaaa", str = "dog cat cat dog" should return false.pattern = "abba", str = "dog dog dog dog" should return false.

Notes:
You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.

Solution:

建立两个map，相当于建立一个双向映射，对每一个pattern建立到一个单词的映射，同时建立一个反映射，然后判断后续的已出现的pattern是否对应正确的单词就好。注意pattern比单词多或单词比pattern多的情况就好。
AC后看了一眼别人的代码，思路是类似的，不过不用建立第二个映射，使用set记录已出现的单词就好，还有获得单词直接使用索引，不用每次都把string的一部分去掉，节省时间。
时间复杂度：O（n）
空间复杂度：O（n）

class Solution {public:    bool wordPattern(string pattern, string str) {        map<char, string> m1;        map<string, char> m2;        for (int i = 0; i < pattern.size(); i++) {            string tmp = getNext(str);            if (tmp == "") return false;            if (m1.find(pattern[i]) == m1.end()) {                if (m2.find(tmp) == m2.end()) {                    m1[pattern[i]] = tmp;                    m2[tmp] = pattern[i];                } else {                    cout << "1 " << tmp;                    return false;                }            } else if (m1[pattern[i]] != tmp) {                return false;            }        }        if (str.empty())            return true;        return false;    }    string getNext(string & str) {        string ret;        for (int i = 0; i < str.size(); i++) {            if (str[i] == ' ') {                str = str.substr(i+1);                return ret;            }            ret += str[i];        }        ret = str;        str = "";        return ret;    }};