68 lines
1.4 KiB
JavaScript
68 lines
1.4 KiB
JavaScript
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/*
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* @lc app=leetcode id=338 lang=javascript
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*
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* [338] Counting Bits
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*
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* https://leetcode.com/problems/counting-bits/description/
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*
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* algorithms
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* Medium (64.04%)
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* Total Accepted: 163.1K
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* Total Submissions: 253K
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* Testcase Example: '2'
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*
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* Given a non negative integer number num. For every numbers i in the range 0
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* ≤ i ≤ num calculate the number of 1's in their binary representation and
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* return them as an array.
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*
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* Example 1:
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*
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*
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* Input: 2
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* Output: [0,1,1]
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*
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* Example 2:
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*
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*
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* Input: 5
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* Output: [0,1,1,2,1,2]
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*
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*
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* Follow up:
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*
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*
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* It is very easy to come up with a solution with run time
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* O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a
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* 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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*/
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/**
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* @param {number} num
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* @return {number[]}
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*/
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var countBits = function(num) {
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// tag: bit dp
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// Time complexity: O(n)
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// Space complexity: O(n)
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const res = [];
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res[0] = 0;
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// 10000100110101
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for (let i = 1; i <= num; i++) {
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if ((i & 1) === 0) {
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// 偶数
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// 偶数最后一位是0，因此右移一位对结果没有影响
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res[i] = res[i >> 1];
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} else {
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// 奇数
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// 奇数最后一位是1，i - 1 的 位数 + 1 就是结果
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res[i] = res[i - 1] + 1;
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}
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}
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return res;
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};
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