LeetCode-779. K-th Symbol in Grammar

On the first row, we write a 0. Now in every subsequent row, we look at the previous row and replace each occurrence of 0 with 01, and each occurrence of 1 with 10.

Given row N and index K, return the K-th indexed symbol in row N. (The values of K are 1-indexed.) (1 indexed).

Examples:
Input: N = 1, K = 1
Output: 0

Input: N = 2, K = 1
Output: 0

Input: N = 2, K = 2
Output: 1

Input: N = 4, K = 5
Output: 1

Explanation:
row 1: 0
row 2: 01
row 3: 0110
row 4: 01101001

Note:

1. N will be an integer in the range [1, 30].
2. K will be an integer in the range [1, 2^(N-1)].

题解：

第一次超时：

class Solution {
public:
    int kthGrammar(int N, int K) {
        vector<int> v(1, 0);
        for (int i = 1; i < N; i++) {
          for (int j = 0; j < pow(2, i); j+= 2) {
            if (v[j] == 0) {
              v.insert(v.begin() + j + 1, 1);
            }
            else if (v[j] == 1) {
              v.insert(v.begin() + j + 1, 0);
            }
          }
        }
        return v[K - 1];
    }
};

第二次内存溢出：

class Solution {
public:
    int kthGrammar(int N, int K) {
        string idx = "0";
        for (int i = 0; i < N - 1; i++) {
          string idx2 = "";
          for (int j = 0; j < idx.length(); j++) {
            if (idx[j] == '0') {
              idx2 += "01";
            }
            else if (idx[j] == '1') {
              idx2 += "10";
            }
          }
          idx.clear();
          idx = idx2;
        }
        return idx[K - 1] - 48;
    }
};

心态崩了哟= =，那肯定不是借助额外内存空间或者遍历来解题了，应该是数学规律，每次的串前一半跟上一串相同，后一半是前一半的取反，ac。

class Solution {
public:
  int kthGrammar(int N, int K) {
    if (N == 1) {
      return 0;
    }
    if (K <= pow(2, N - 2)) {
      return kthGrammar(N - 1, K);
    }
    else {
      if (kthGrammar(N - 1, K - pow(2, N - 2)) == 0) {
        return 1;
      }
      else {
        return 0;
      }
    }
  }
};