leetcode 600 Count number of binary strings without consecutive 1

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 functionsint main(){    cout << countStrings(3) << endl;    return 0;}