给定两个整数n和k，计算长度为n的二进制字符串的数量，其中k为相邻1出现的次数。
例子：
Input : n = 5, k = 2
Output : 6
Explanation:
Binary strings of length 5 in which k number of times
two adjacent set bits appear.
00111
01110
11100
11011
10111
11101
Input : n = 4, k = 1
Output : 3
Explanation:
Binary strings of length 3 in which k number of times
two adjacent set bits appear.
0011
1100
0110
让我们尝试为上述问题语句编写递归函数：
1)n = 1，仅存在两个长度为1的二进制字符串，而没有相邻的1
字符串1：“ 0”
字符串2：“ 1”
2)对于所有n> 1和所有k，出现两种情况
a)以0结尾的字符串：长度为n的字符串可以通过将长度n-1的所有字符串附加0到k乘以两个相邻的1以0和1结尾的相邻字符串来创建(在第n个位置为0不会更改计数)的相邻1个)。
b)以1结尾的字符串：可以通过将1附加到所有k乘以1并相邻的长度为n-1的字符串，并以0结尾以及所有k-1邻近于1的n-1的所有字符串而将n附加为n，来创建长度为n的字符串。以1结尾。
示例：let s = 011，即一个以1结尾的字符串，其相邻计数为1。将1加s = 0111会增加相邻1的计数。
Let there be an array dp[i][j][2] where dp[i][j][0]
denotes number of binary strings with length i having
j number of two adjacent 1's and ending with 0.
Similarly dp[i][j][1] denotes the same binary strings
with length i and j adjacent 1's but ending with 1.
Then:
dp[1][0][0] = 1 and dp[1][0][1] = 1
For all other i and j,
dp[i][j][0] = dp[i-1][j][0] + dp[i-1][j][1]
dp[i][j][1] = dp[i-1][j][0] + dp[i-1][j-1][1]
Then, output dp[n][k][0] + dp[n][k][1]
C++
// C++ program to count number of binary strings
// with k times appearing consecutive 1's.
#include
using namespace std;
int countStrings(int n, int k)
{
// dp[i][j][0] stores count of binary
// strings of length i with j consecutive
// 1's and ending at 0.
// dp[i][j][1] stores count of binary
// strings of length i with j consecutive
// 1's and ending at 1.
int dp[n + 1][k + 1][2];
memset(dp, 0, sizeof(dp));
// If n = 1 and k = 0.
dp[1][0][0] = 1;
dp[1][0][1] = 1;
for (int i = 2; i <= n; i++) {
// number of adjacent 1's can not exceed i-1
for (int j = 0; j <= k; j++) {
dp[i][j][0] = dp[i - 1][j][0] + dp[i - 1][j][1];
dp[i][j][1] = dp[i - 1][j][0];
if (j - 1 >= 0)
dp[i][j][1] += dp[i - 1][j - 1][1];
}
}
return dp[n][k][0] + dp[n][k][1];
}
// Driver code
int main()
{
int n = 5, k = 2;
cout << countStrings(n, k);
return 0;
} Java
// Java program to count number of binary strings
// with k times appearing consecutive 1's.
class GFG {
static int countStrings(int n, int k)
{
// dp[i][j][0] stores count of binary
// strings of length i with j consecutive
// 1's and ending at 0.
// dp[i][j][1] stores count of binary
// strings of length i with j consecutive
// 1's and ending at 1.
int dp[][][] = new int[n + 1][k + 1][2];
// If n = 1 and k = 0.
dp[1][0][0] = 1;
dp[1][0][1] = 1;
for (int i = 2; i <= n; i++) {
// number of adjacent 1's can not exceed i-1
for (int j = 0; j < i && j < k + 1; j++) {
dp[i][j][0] = dp[i - 1][j][0] + dp[i - 1][j][1];
dp[i][j][1] = dp[i - 1][j][0];
if (j - 1 >= 0) {
dp[i][j][1] += dp[i - 1][j - 1][1];
}
}
}
return dp[n][k][0] + dp[n][k][1];
}
// Driver code
public static void main(String[] args)
{
int n = 5, k = 2;
System.out.println(countStrings(n, k));
}
}
// This code has been contributed by 29AjayKumarPython3
# Python3 program to count number of
# binary strings with k times appearing
# consecutive 1's.
def countStrings(n, k):
# dp[i][j][0] stores count of binary
# strings of length i with j consecutive
# 1's and ending at 0.
# dp[i][j][1] stores count of binary
# strings of length i with j consecutive
# 1's and ending at 1.
dp = [[[0, 0] for __ in range(k + 1)]
for _ in range(n + 1)]
# If n = 1 and k = 0.
dp[1][0][0] = 1
dp[1][0][1] = 1
for i in range(2, n + 1):
# number of adjacent 1's can not exceed i-1
for j in range(k + 1):
dp[i][j][0] = (dp[i - 1][j][0] +
dp[i - 1][j][1])
dp[i][j][1] = dp[i - 1][j][0]
if j >= 1:
dp[i][j][1] += dp[i - 1][j - 1][1]
return dp[n][k][0] + dp[n][k][1]
# Driver Code
if __name__ == '__main__':
n = 5
k = 2
print(countStrings(n, k))
# This code is contributed by vibhu4agarwalC#
// C# program to count number of binary strings
// with k times appearing consecutive 1's.
using System;
class GFG {
static int countStrings(int n, int k)
{
// dp[i][j][0] stores count of binary
// strings of length i with j consecutive
// 1's and ending at 0.
// dp[i][j][1] stores count of binary
// strings of length i with j consecutive
// 1's and ending at 1.
int[,, ] dp = new int[n + 1, k + 1, 2];
// If n = 1 and k = 0.
dp[1, 0, 0] = 1;
dp[1, 0, 1] = 1;
for (int i = 2; i <= n; i++) {
// number of adjacent 1's can not exceed i-1
for (int j = 0; j < i && j < k + 1; j++) {
dp[i, j, 0] = dp[i - 1, j, 0] + dp[i - 1, j, 1];
dp[i, j, 1] = dp[i - 1, j, 0];
if (j - 1 >= 0) {
dp[i, j, 1] += dp[i - 1, j - 1, 1];
}
}
}
return dp[n, k, 0] + dp[n, k, 1];
}
// Driver code
public static void Main(String[] args)
{
int n = 5, k = 2;
Console.WriteLine(countStrings(n, k));
}
}
// This code contributed by Rajput-JiPHP
= 0 && isset($dp[$i][$j][1]))
$dp[$i][$j][1] += $dp[$i - 1][$j - 1][1];
}
}
return $dp[$n][$k][0] + $dp[$n][$k][1];
}
// Driver code
$n=5;
$k=2;
echo countStrings($n, $k);
// This code is contributed by mits
?>输出：
6
时间复杂度： O(n 2 )