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  • LeetCode——Counting Bits

    Question

    Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

    Example:
    For num = 5 you should return [0,1,1,2,1,2].

    Follow up:

    It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
    Space complexity should be O(n).
    Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.
    Credits:
    Special thanks to @ syedee for adding this problem and creating all test cases.

    Solution

    动态规划,这里要应用求一个整数,对应二进制数中1的个数的方法,i & (i - 1),也就是将i的最右边的1置为0.

    那么每个整数中1的个数,就等于不包含最右边1以后1的个数再加1。 ret[i] = ret[i & (i - 1)] + 1.

    Code

    class Solution {
    public:
        vector<int> countBits(int num) {
            vector<int> table(num + 1, 0);
            for (int i = 1; i <= num; i++) {
                table[i] = table[i & (i - 1)] + 1;
            }
            return table;
        }
    };
    
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  • 原文地址:https://www.cnblogs.com/zhonghuasong/p/7501834.html
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