Distinct binary strings of length n with no consecutive 1s

最新推荐文章于 2024-06-23 01:44:59 发布

原创最新推荐文章于 2024-06-23 01:44:59 发布 · 536 阅读

0 ·

CC 4.0 BY-SA版权

Google 专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一种算法，用于计算长度为n的不含连续1的不同二进制字符串的数量。通过递推方法，分别计算以0和1结尾的最大数量，最终得出总数量。

Count all possible distinct binary strings of length n with no consecutive 1s

1: Get the max number ending with 0, Get the max number ending with 1

2: For the next max number ending with 0, we just need to append 0 to prev_max_ending_zero + append 0 to prev_max_ending_one

3: For the next max number ending with 1, we can only append 1 to prev_max_ending_zero.

#include <iostream>
#include <vector>
using namespace std;

int totalNumber(int n) {
  // ends with 0
  int zn = 1;

  // ends with 1
  int cn = 0;
  int total = 0;
  for(int i = 1; i <= n; ++i) {
    int tmp = zn;
    zn = zn + cn;
    cn = tmp;
    total = zn + cn;
  }
  return total;
}

int main(void) {
  cout << totalNumber(4) << endl;
}