[Leetcode] 779. K-th Symbol in Grammar 解题报告_on the first row, we write a 0, now in-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/79707453

本文介绍了一种递归构造语法表的方法，通过该方法可以确定第N行第K个符号的具体值。文中提供了详细的实现思路及C++代码示例。

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题目：

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:

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

思路：

第N行的第K个数需要根据第N-1行的第(K - 1) / 2 +1个数来确定。所以我们首先递归求解上一行中对应位置上的数，然后再推导出当前行上K位置上的数即可。实现上的小技巧就是0-indexed表示和1-indexed表示之间的相互转化。

代码：

class Solution {
public:
    int kthGrammar(int N, int K) {
        if (N == 1) {
            return 0;
        }
        int last_K = (K - 1) / 2;                   // convert to 0-indexed
        int remain = (K - 1) % 2;                   // convert to 0-indexed
        int ans = kthGrammar(N - 1, last_K + 1);    // convert back to 1-indexed
        if (ans == 0) {
            return remain == 0 ? 0 : 1;
        }
        else {
            return remain == 0 ? 1 : 0;
        }
    }
};