600. Non-negative Integers without Consecutive Ones

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Input: 5
Output: 5
Explanation: 
Here are the non-negative integers <= 5 with their corresponding binary representations:
0 : 0
1 : 1
2 : 10
3 : 11
4 : 100
5 : 101
Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule. 

Approach #1: DP. [C++]

class Solution {
public:
    int findIntegers(int num) {
        vector<int> f(35, 0);
        f[0] = 1;
        f[1] = 2;
        for (int i = 2; i < 32; ++i) 
            f[i] = f[i-1] + f[i-2];
        int ans = 0, k = 30, pre_bit = 0;
        while (k >= 0) {
            if (num & (1 << k)) {
                ans += f[k];
                if (pre_bit == 1) return ans;
                pre_bit = 1;
            } else pre_bit = 0;
            k--;
        }
        return ans+1;
    }
};

Analysis:

The solution if based on 2 fact:

First: the number of length k string without consecutive 1 is Fibonacci sequence f(k);

For example, is k = 5, the range is 00000 - 11111. We can consider it as two ranges, which are 00000 - 01111 ans 10000  10111. any number >= 11000 is not allowed due to consecutive 1. The first case is actually f(4), and the second case is f(3), so f(5) = f(4) + f(3).

Second: Scan the number from most significant digit, i.e. left to right, in binary format. If we find a ‘1‘ with k digits to the right, count increases by f(k) beause we can put a ‘0‘ at this digit and any valid length k string behind; After that, we continue the loop to consider the remaining case, i.e. we put a ‘1‘ at this digit. If consecutive 1s are found, we exit the loop and return the answer. By the end of the loop, we return ans + 1 to include the number n itself.

Reference:

https://leetcode.com/problems/non-negative-integers-without-consecutive-ones/discuss/103754/C%2B%2B-Non-DP-O(32)-Fibonacci-solution

600. Non-negative Integers without Consecutive Ones

标签：his   --   discuss   class   ber   NPU   number   hat   exit   

原文地址：https://www.cnblogs.com/ruruozhenhao/p/10503426.html