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124 changes: 75 additions & 49 deletions solution/0200-0299/0278.First Bad Version/README.md
Original file line number Diff line number Diff line change
Expand Up @@ -30,8 +30,8 @@ tags:
<strong>输入:</strong>n = 5, bad = 4
<strong>输出:</strong>4
<strong>解释:</strong>
<code>调用 isBadVersion(3) -&gt; false
调用 isBadVersion(5)&nbsp;-&gt; true
<code>调用 isBadVersion(3) -&gt; false
调用 isBadVersion(5)&nbsp;-&gt; true
调用 isBadVersion(4)&nbsp;-&gt; true</code>
<code>所以,4 是第一个错误的版本。</code>
</pre>
Expand All @@ -57,33 +57,35 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一:二分查找

我们定义二分查找的左边界 $l = 1$,右边界 $r = n$。

当 $l < r$ 时,我们计算中间位置 $\textit{mid} = \left\lfloor \frac{l + r}{2} \right\rfloor$,然后调用 `isBadVersion(mid)` 接口,如果返回 $\textit{true}$,则说明第一个错误的版本在 $[l, \textit{mid}]$ 之间,我们令 $r = \textit{mid}$;否则第一个错误的版本在 $[\textit{mid} + 1, r]$ 之间,我们令 $l = \textit{mid} + 1$。

最终返回 $l$ 即可。

时间复杂度 $O(\log n)$,空间复杂度 $O(1)$。

<!-- tabs:start -->

#### Python3

```python
# The isBadVersion API is already defined for you.
# @param version, an integer
# @return an integer
# def isBadVersion(version):
# def isBadVersion(version: int) -> bool:


class Solution:
def firstBadVersion(self, n):
"""
:type n: int
:rtype: int
"""
left, right = 1, n
while left < right:
mid = (left + right) >> 1
def firstBadVersion(self, n: int) -> int:
l, r = 1, n
while l < r:
mid = (l + r) >> 1
if isBadVersion(mid):
right = mid
r = mid
else:
left = mid + 1
return left
l = mid + 1
return l
```

#### Java
Expand All @@ -94,16 +96,16 @@ class Solution:

public class Solution extends VersionControl {
public int firstBadVersion(int n) {
int left = 1, right = n;
while (left < right) {
int mid = (left + right) >>> 1;
int l = 1, r = n;
while (l < r) {
int mid = (l + r) >>> 1;
if (isBadVersion(mid)) {
right = mid;
r = mid;
} else {
left = mid + 1;
l = mid + 1;
}
}
return left;
return l;
}
}
```
Expand All @@ -117,16 +119,16 @@ public class Solution extends VersionControl {
class Solution {
public:
int firstBadVersion(int n) {
int left = 1, right = n;
while (left < right) {
int mid = left + ((right - left) >> 1);
int l = 1, r = n;
while (l < r) {
int mid = l + (r - l) / 2;
if (isBadVersion(mid)) {
right = mid;
r = mid;
} else {
left = mid + 1;
l = mid + 1;
}
}
return left;
return l;
}
};
```
Expand All @@ -143,19 +145,45 @@ public:
*/

func firstBadVersion(n int) int {
left, right := 1, n
for left < right {
mid := (left + right) >> 1
l, r := 1, n
for l < r {
mid := (l + r) >> 1
if isBadVersion(mid) {
right = mid
r = mid
} else {
left = mid + 1
l = mid + 1
}
}
return left
return l
}
```

#### TypeScript

```ts
/**
* The knows API is defined in the parent class Relation.
* isBadVersion(version: number): boolean {
* ...
* };
*/

var solution = function (isBadVersion: any) {
return function (n: number): number {
let [l, r] = [1, n];
while (l < r) {
const mid = (l + r) >>> 1;
if (isBadVersion(mid)) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
};
```

#### Rust

```rust
Expand All @@ -165,17 +193,16 @@ func firstBadVersion(n int) int {

impl Solution {
pub fn first_bad_version(&self, n: i32) -> i32 {
let mut left = 1;
let mut right = n;
while left < right {
let mid = left + (right - left) / 2;
let (mut l, mut r) = (1, n);
while l < r {
let mid = l + (r - l) / 2;
if self.isBadVersion(mid) {
right = mid;
r = mid;
} else {
left = mid + 1;
l = mid + 1;
}
}
left
l
}
}
```
Expand Down Expand Up @@ -203,17 +230,16 @@ var solution = function (isBadVersion) {
* @return {integer} The first bad version
*/
return function (n) {
let left = 1;
let right = n;
while (left < right) {
const mid = (left + right) >>> 1;
let [l, r] = [1, n];
while (l < r) {
const mid = (l + r) >>> 1;
if (isBadVersion(mid)) {
right = mid;
r = mid;
} else {
left = mid + 1;
l = mid + 1;
}
}
return left;
return l;
};
};
```
Expand Down
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