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lines changed Original file line number Diff line number Diff line change 65
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<!-- 这里可写通用的实现逻辑 -->
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- ` n & (n - 1) ` 会消除 n 中最后一位中的 1。
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+ 朴素解法:一位一位数。
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+
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+ ``` txt
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+ HAMMING-WEIGHT(n)
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+ r = 0
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+ while n != 0
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+ r += n & 1
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+ n >>= 1
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+ r
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+ ```
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+
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+ 利用 ` n & (n - 1) ` 消除 ` n ` 中最后一位 1 这一特点,优化过程:
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+
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+ ``` txt
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+ HAMMING-WEIGHT(n)
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+ r = 0
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+ while n != 0
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+ n &= n - 1
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+ r += 1
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+ r
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+ ```
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+
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+ 以 5 为例,演示推演过程:
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+
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+ ``` txt
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+ [0, 1, 0, 1] // 5
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+ [0, 1, 0, 0] // 5 - 1 = 4
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+ [0, 1, 0, 0] // 5 & 4 = 4
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+
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+ [0, 1, 0, 0] // 4
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+ [0, 0, 1, 1] // 4 - 1 = 3
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+ [0, 0, 0, 0] // 4 & 3 = 0
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+ ```
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同 [ 剑指 Offer 15. 二进制中 1 的个数] ( /lcof/面试题15.%20二进制中1的个数/README.md )
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@@ -150,6 +182,42 @@ var hammingWeight = function (n) {
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};
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```
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+ ### ** Rust**
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+
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+ ``` rust
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+ impl Solution {
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+ pub fn hammingWeight (n : u32 ) -> i32 {
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+ n . count_ones () as i32
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+ }
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+ }
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+ ```
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+
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+ ``` rust
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+ impl Solution {
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+ pub fn hammingWeight (mut n : u32 ) -> i32 {
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+ let mut res = 0 ;
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+ while n != 0 {
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+ res += n & 1 ;
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+ n >>= 1 ;
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+ }
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+ res as i32
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+ }
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+ }
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+ ```
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+
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+ ``` rust
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+ impl Solution {
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+ pub fn hammingWeight (mut n : u32 ) -> i32 {
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+ let mut res = 0 ;
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+ while n != 0 {
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+ n &= (n - 1 );
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+ res += 1 ;
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+ }
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+ res
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+ }
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+ }
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+ ```
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+
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### ** ...**
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```
Original file line number Diff line number Diff line change @@ -127,6 +127,42 @@ var hammingWeight = function (n) {
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};
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```
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+ ### ** Rust**
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+
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+ ``` rust
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+ impl Solution {
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+ pub fn hammingWeight (n : u32 ) -> i32 {
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+ n . count_ones () as i32
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+ }
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+ }
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+ ```
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+
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+ ``` rust
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+ impl Solution {
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+ pub fn hammingWeight (mut n : u32 ) -> i32 {
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+ let mut res = 0 ;
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+ while n != 0 {
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+ res += n & 1 ;
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+ n >>= 1 ;
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+ }
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+ res as i32
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+ }
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+ }
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+ ```
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+
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+ ``` rust
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+ impl Solution {
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+ pub fn hammingWeight (mut n : u32 ) -> i32 {
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+ let mut res = 0 ;
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+ while n != 0 {
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+ n &= (n - 1 );
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+ res += 1 ;
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+ }
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+ res
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+ }
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+ }
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+ ```
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+
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### ** ...**
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```
Original file line number Diff line number Diff line change
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+ impl Solution {
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+ pub fn hammingWeight ( n : u32 ) -> i32 {
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+ n. count_ones ( ) as i32
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+ }
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+ }
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