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  • 600. Non-negative Integers without Consecutive Ones

    Given a positive integer n, find the number of non-negative integers less than or equal to n, whose binary representations do NOT contain consecutive ones.

    Example 1:

    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. 
    

    Note: 1 <= n <= 109

     

    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

    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/10503426.html
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