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Commit 37aa93a

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feat: add rust solution to lc problem: No.0191 (doocs#784)
No.0191.Number of 1 Bits
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‎solution/0100-0199/0191.Number of 1 Bits/README.md‎

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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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```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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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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```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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以 5 为例,演示推演过程:
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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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[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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```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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```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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```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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```

‎solution/0100-0199/0191.Number of 1 Bits/README_EN.md‎

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@@ -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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```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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```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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```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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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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