Commit 385b906

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feat: add solutions to lc problem: No.1611 (#859)

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Lines changed: 34 additions & 4 deletions

Original file line number	Diff line number	Diff line change
`@@ -51,22 +51,52 @@`
`51`	`51`
`52`	`52`	`<!-- 这里可写通用的实现逻辑 -->`
`53`	`53`
	`54`	`+方法一:递归`
	`55`	`+`
	`56`	`+通过找规律可以发现动态规划转移方程如下:`
	`57`	`+`
	`58`	`+$$`
	`59`	`+dp[n] = dp[2^k] - dp[n - 2^k]`
	`60`	`+$$`
	`61`	`+`
	`62`	`+其中 $dp[2^k] = 2^{k+1}-1,ドル而 $k$ 表示小于等于 $n$ 的最大的 2ドル$ 的整数次幂的位数,即 2ドル^k$ 是小于等于 $n$ 的最大的 2ドル$ 的整数次幂。`
	`63`	`+`
`54`	`64`	`<!-- tabs:start -->`
`55`	`65`
`56`	`66`	`### Python3`
`57`	`67`
`58`	`68`	`<!-- 这里可写当前语言的特殊实现逻辑 -->`
`59`	`69`
`60`	`70`	```python
`61`		`-`
	`71`	`+class Solution:`
	`72`	`+ def minimumOneBitOperations(self, n: int) -> int:`
	`73`	`+ if n <= 1:`
	`74`	`+ return n`
	`75`	`+ for i in range(64):`
	`76`	`+ if (n >> i) == 1:`
	`77`	`+ base = 1 << i`
	`78`	`+ break`
	`79`	`+ return 2*base-1 - self.minimumOneBitOperations(n-base)`
`62`	`80`	```
`63`	`81`
`64`		`-### Java`
	`82`	`+### Go`
`65`	`83`
`66`	`84`	`<!-- 这里可写当前语言的特殊实现逻辑 -->`
`67`	`85`
`68`		-```java
`69`		`-`
	`86`	+```go
	`87`	`+func minimumOneBitOperations(n int) int {`
	`88`	`+ if n <= 1 {`
	`89`	`+ return n`
	`90`	`+ }`
	`91`	`+ base := 0`
	`92`	`+ for i := 0; i < 64; i++ {`
	`93`	`+ if (n >> i) == 1 {`
	`94`	`+ base = 1 << i`
	`95`	`+ break`
	`96`	`+ }`
	`97`	`+ }`
	`98`	`+ return (base << 1) - 1 - minimumOneBitOperations(n-base)`
	`99`	`+}`
`70`	`100`	```
`71`	`101`
`72`	`102`	`### ...`

Lines changed: 25 additions & 5 deletions

Original file line number	Diff line number	Diff line change
`@@ -50,13 +50,33 @@`
`50`	`50`	`### Python3`
`51`	`51`
`52`	`52`	```python
`53`		`-`
	`53`	`+class Solution:`
	`54`	`+ def minimumOneBitOperations(self, n: int) -> int:`
	`55`	`+ if n <= 1:`
	`56`	`+ return n`
	`57`	`+ for i in range(64):`
	`58`	`+ if (n >> i) == 1:`
	`59`	`+ base = 1 << i`
	`60`	`+ break`
	`61`	`+ return 2*base-1 - self.minimumOneBitOperations(n-base)`
`54`	`62`	```
`55`	`63`
`56`		`-### Java`
`57`		`-`
`58`		-```java
`59`		`-`
	`64`	`+### Go`
	`65`	`+`
	`66`	+```go
	`67`	`+func minimumOneBitOperations(n int) int {`
	`68`	`+ if n <= 1 {`
	`69`	`+ return n`
	`70`	`+ }`
	`71`	`+ base := 0`
	`72`	`+ for i := 0; i < 64; i++ {`
	`73`	`+ if (n >> i) == 1 {`
	`74`	`+ base = 1 << i`
	`75`	`+ break`
	`76`	`+ }`
	`77`	`+ }`
	`78`	`+ return (base << 1) - 1 - minimumOneBitOperations(n-base)`
	`79`	`+}`
`60`	`80`	```
`61`	`81`
`62`	`82`	`### ...`

Lines changed: 13 additions & 0 deletions

Original file line number	Diff line number	Diff line change
`@@ -0,0 +1,13 @@`
	`1`	`+func minimumOneBitOperations(n int) int {`
	`2`	`+ if n <= 1 {`
	`3`	`+ return n`
	`4`	`+ }`
	`5`	`+ base := 0`
	`6`	`+ for i := 0; i < 64; i++ {`
	`7`	`+ if (n >> i) == 1 {`
	`8`	`+ base = 1 << i`
	`9`	`+ break`
	`10`	`+ }`
	`11`	`+ }`
	`12`	`+ return (base << 1) - 1 - minimumOneBitOperations(n-base)`
	`13`	`+}`

Lines changed: 9 additions & 0 deletions

Original file line number	Diff line number	Diff line change
`@@ -0,0 +1,9 @@`
	`1`	`+class Solution:`
	`2`	`+ def minimumOneBitOperations(self, n: int) -> int:`
	`3`	`+ if n <= 1:`
	`4`	`+ return n`
	`5`	`+ for i in range(64):`
	`6`	`+ if (n >> i) == 1:`
	`7`	`+ base = 1 << i`
	`8`	`+ break`
	`9`	`+ return 2*base-1 - self.minimumOneBitOperations(n-base)`

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