68 lines
2.4 KiB
C++
68 lines
2.4 KiB
C++
// Source : https://leetcode.com/problems/counting-bits/
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// Author : Hao Chen
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// Date : 2016-05-30
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/***************************************************************************************
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*
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* Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤
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* num calculate the number of 1's in their binary representation and return them as an
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* array.
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*
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* Example:
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* For num = 5 you should return [0,1,1,2,1,2].
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*
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* Follow up:
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*
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* It is very easy to come up with a solution with run time O(n*sizeof(integer)). But
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* can you do it in linear time O(n) /possibly in a single pass?
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* Space complexity should be O(n).
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* Can you do it like a boss? Do it without using any builtin function like
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* __builtin_popcount in c++ or in any other language.
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*
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* You should make use of what you have produced already.
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* Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to
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* generate new range from previous.
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* Or does the odd/even status of the number help you in calculating the number of 1s?
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*
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* Credits:Special thanks to @ syedee for adding this problem and creating all test
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* cases.
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***************************************************************************************/
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class Solution {
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public:
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/*
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* Initialization
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*
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* bits_cnt[0] => 0000 => 0
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* bits_cnt[1] => 0001 => 1
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*
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* if the number has 2 bits (2, 3), then we can split the binary to two parts,
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* 2 = 10 + 0 and 3= 10 + 1, then we can reuse the bits_cnt[0] and bits_cnt[1]
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*
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* bits_cnt[2] => 0010 => 0010 + 0 => 1 + bits_cnt[0];
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* bits_cnt[3] => 0011 => 0010 + 1 => 1 + bits_cnt[1];
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*
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* if the number has 3 bits (4,5,6,7), then we can split the binary to two parts,
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* 4 = 100 + 0, 5 = 100 + 01, 6= 100 + 10, 7 = 100 +11
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* then we can reuse the bits_cnt[0] and bits_cnt[1]
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*
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* bits_cnt[4] => 0110 => 0100 + 00 => 1 + bits_cnt[0];
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* bits_cnt[5] => 0101 => 0100 + 01 => 1 + bits_cnt[1];
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* bits_cnt[6] => 0110 => 0100 + 10 => 1 + bits_cnt[2];
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* bits_cnt[7] => 0111 => 0100 + 11 => 1 + bits_cnt[3];
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*
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* so, we can have the solution:
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*
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* bits_cnt[x] = bits_cnt[x & (x-1) ] + 1;
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*
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*/
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vector<int> countBits(int num) {
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vector<int> bits_cnt(num+1, 0);
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for (int i=1; i<=num; i++) {
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bits_cnt[i] = bits_cnt[i & (i-1)] + 1;
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}
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return bits_cnt;
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}
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};
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