leetcode 600 Count number of binary strings without consecutive 1 不出现连续1二进制数量

Given a positive integer N, count all possible distinct binary strings of length N such that there are no consecutive 1’s.

Examples:

Input: N = 2

Output: 3

// The 3 strings are 00, 01, 10

Input: N = 3

Output: 5

// The 5 strings are 000, 001, 010, 100, 101

思路是用DP，考虑从n位增加一位到n+1位的情况。如果首位是0，那么第n+1位的首位既可以是0，也可以是1，因为都不会产生连续的1；如果首位是1，那么只能生成首位是0的n+1位数，才能避免出现连续的1。

用zeros记录首

代码如下：

// C++ program to count all distinct binary strings
// without two consecutive 1's
#include <iostream>
using namespace std;

int countStrings(const int n)
{
vector<int> a(n), b(n);
a[0] = b[0] = 1;
for (int i = 1; i < n; i++)
{
a[i] = a[i - 1] + b[i - 1];
b[i] = a[i - 1];
}
return a[n - 1] + b[n - 1];
}

// Driver program to test above functions
int main()
{
cout << countStrings(3) << endl;
return 0;
}