1415. The k-th Lexicographical String of All Happy Strings of Length n

问题：

求由n个字符构成，按照字母序排序后，第k个happy string。

happy string定义：

• consists only of letters of the set ['a', 'b', 'c'].
  • 仅由a，b，c构成
• s[i] != s[i + 1] for all values of i from 1 to s.length - 1 (string is 1-indexed).
  • 连续两个字符不重复

如："abc", "ac", "b", "abcbabcbcb"

不是happy string的例子："aa", "baa", "ababbc" 

Example 1:
Input: n = 1, k = 3
Output: "c"
Explanation: The list ["a", "b", "c"] contains all happy strings of length 1. The third string is "c".

Example 2:
Input: n = 1, k = 4
Output: ""
Explanation: There are only 3 happy strings of length 1.

Example 3:
Input: n = 3, k = 9
Output: "cab"
Explanation: There are 12 different happy string of length 3 ["aba", "abc", "aca", "acb", "bab", "bac", "bca", "bcb", "cab", "cac", "cba", "cbc"]. You will find the 9th string = "cab"

Example 4:
Input: n = 2, k = 7
Output: ""

Example 5:
Input: n = 10, k = 100
Output: "abacbabacb"

Constraints:
1 <= n <= 10
1 <= k <= 100

　　

解法：Backtracking（回溯算法）

• 状态：到当前位置pos为止，构成的快乐串path
• 选择：a，b，c中，除了path.back字符之外的两个。
• 退出递归条件：
  • path.length==n ->处理k--（找到一个快乐串）
  • 得到要求的第k个结果，条件：k==0

代码参考：

class Solution {
public:
    void backtrack(string& res, int pos, string path, int& n, int& k) {
        if(path.length() == n){
            k--;
            if(k==0){
                res=path;
            }
            return;
        }
        for(int i=0; i<3; i++) {
            char cur = 'a'+i;
            if('a'+i==path.back()) continue;
            backtrack(res, pos+1, path+cur, n, k);
            if(k==0) return;
        }
        return;
    }
    string getHappyString(int n, int k) {
        string res("");
        backtrack(res, 0, string(""), n, k);
        return res;
    }
};