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Commit 8f504fc

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‎problems/0338.counting-bits/README.md

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# 0338.counting-bits
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```text
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338. 比特位计数
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给定一个非负整数 num。对于 0 ≤ i ≤ num 范围中的每个数字 i ,计算其二进制数中的 1 的数目并将它们作为数组返回。
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示例 1:
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输入: 2
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输出: [0,1,1]
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示例 2:
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输入: 5
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输出: [0,1,1,2,1,2]
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进阶:
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给出时间复杂度为O(n*sizeof(integer))的解答非常容易。但你可以在线性时间O(n)内用一趟扫描做到吗?
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要求算法的空间复杂度为O(n)。
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你能进一步完善解法吗?要求在C++或任何其他语言中不使用任何内置函数(如 C++ 中的 __builtin_popcount)来执行此操作。
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来源:力扣(LeetCode)
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链接:https://leetcode-cn.com/problems/counting-bits
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著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
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```

‎problems/0338.counting-bits/run.go

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package bits
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import "fmt"
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func Run() {
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bits := countBits(5)
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fmt.Println(bits)
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}
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func countBits(num int) []int {
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bits := make([]int, num+1)
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for i := 0; i <= num; i++ {
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bits[i] = oneCount(i)
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}
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return bits
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}
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func oneCount(n int) int {
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c := 0
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for ; n > 0; n &= n - 1 {
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c++
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}
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return c
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}

‎problems/0338.counting-bits/run_test.go

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package bits
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import "testing"
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func TestRun(t *testing.T) {
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Run()
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}

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