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;
}
}
};