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Merged
yanglbme merged 1 commit into from
Apr 20, 2026
Merged

feat: add solutions for lc No.2078 #5168

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153 changes: 62 additions & 91 deletions solution/2000-2099/2078.Two Furthest Houses With Different Colors/README.md
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -78,89 +78,13 @@ tags:

<!-- solution:start -->

### 方法一：暴力枚举
### 方法一：贪心

时间复杂度 $O(n^2)$。
我们可以发现，如果第一栋房子和最后一栋房子颜色不同，那么它们之间的距离就是最大的距离，即 $n - 1$。

<!-- tabs:start -->

#### Python3

```python
class Solution:
def maxDistance(self, colors: List[int]) -> int:
ans, n = 0, len(colors)
for i in range(n):
for j in range(i + 1, n):
if colors[i] != colors[j]:
ans = max(ans, abs(i - j))
return ans
```

#### Java

```java
class Solution {
public int maxDistance(int[] colors) {
int ans = 0, n = colors.length;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (colors[i] != colors[j]) {
ans = Math.max(ans, Math.abs(i - j));
}
}
}
return ans;
}
}
```

#### C++
如果第一栋房子和最后一栋房子颜色相同，那么我们可以从左向右找到第一栋颜色不同的房子，记为 $i$，从右向左找到第一栋颜色不同的房子，记为 $j$，那么最大的距离就是 $\max(n - i - 1, j)$。

```cpp
class Solution {
public:
int maxDistance(vector<int>& colors) {
int ans = 0, n = colors.size();
for (int i = 0; i < n; ++i)
for (int j = i + 1; j < n; ++j)
if (colors[i] != colors[j])
ans = max(ans, abs(i - j));
return ans;
}
};
```

#### Go

```go
func maxDistance(colors []int) int {
ans, n := 0, len(colors)
for i := 0; i < n; i++ {
for j := i + 1; j < n; j++ {
if colors[i] != colors[j] {
ans = max(ans, abs(i-j))
}
}
}
return ans
}

func abs(x int) int {
if x >= 0 {
return x
}
return -x
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### 方法二：贪心
时间复杂度 $O(n)$，其中 $n$ 是房子的数量。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -189,11 +113,13 @@ class Solution {
if (colors[0] != colors[n - 1]) {
return n - 1;
}
int i = 0, j = n - 1;
while (colors[++i] == colors[0])
;
while (colors[--j] == colors[0])
;
int i = 1, j = n - 2;
while (colors[i] == colors[0]) {
++i;
}
while (colors[j] == colors[0]) {
--j;
}
return Math.max(n - i - 1, j);
}
}
Expand All @@ -206,12 +132,16 @@ class Solution {
public:
int maxDistance(vector<int>& colors) {
int n = colors.size();
if (colors[0] != colors[n - 1]) return n - 1;
int i = 0, j = n;
while (colors[++i] == colors[0])
;
while (colors[--j] == colors[0])
;
if (colors[0] != colors[n - 1]) {
return n - 1;
}
int i = 1, j = n - 2;
while (colors[i] == colors[0]) {
++i;
}
while (colors[j] == colors[0]) {
--j;
}
return max(n - i - 1, j);
}
};
Expand All @@ -236,6 +166,47 @@ func maxDistance(colors []int) int {
}
```

#### TypeScript

```ts
function maxDistance(colors: number[]): number {
const n = colors.length;
if (colors[0] !== colors[n - 1]) {
return n - 1;
}
let [i, j] = [1, n - 2];
while (colors[i] === colors[0]) {
i++;
}
while (colors[j] === colors[0]) {
j--;
}
return Math.max(n - i - 1, j);
};
```

#### Rust

```rust
impl Solution {
pub fn max_distance(colors: Vec<i32>) -> i32 {
let n = colors.len();
if colors[0] != colors[n - 1] {
return (n - 1) as i32;
}
let mut i = 1;
while colors[i] == colors[0] {
i += 1;
}
let mut j = n - 2;
while colors[j] == colors[0] {
j -= 1;
}
std::cmp::max((n - i - 1) as i32, j as i32)
}
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
151 changes: 62 additions & 89 deletions solution/2000-2099/2078.Two Furthest Houses With Different Colors/README_EN.md
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -72,87 +72,13 @@ House 0 has color 0, and house 1 has color 1. The distance between them is abs(0

<!-- solution:start -->

### Solution 1
### Solution 1: Greedy

<!-- tabs:start -->

#### Python3

```python
class Solution:
def maxDistance(self, colors: List[int]) -> int:
ans, n = 0, len(colors)
for i in range(n):
for j in range(i + 1, n):
if colors[i] != colors[j]:
ans = max(ans, abs(i - j))
return ans
```

#### Java

```java
class Solution {
public int maxDistance(int[] colors) {
int ans = 0, n = colors.length;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (colors[i] != colors[j]) {
ans = Math.max(ans, Math.abs(i - j));
}
}
}
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
int maxDistance(vector<int>& colors) {
int ans = 0, n = colors.size();
for (int i = 0; i < n; ++i)
for (int j = i + 1; j < n; ++j)
if (colors[i] != colors[j])
ans = max(ans, abs(i - j));
return ans;
}
};
```

#### Go

```go
func maxDistance(colors []int) int {
ans, n := 0, len(colors)
for i := 0; i < n; i++ {
for j := i + 1; j < n; j++ {
if colors[i] != colors[j] {
ans = max(ans, abs(i-j))
}
}
}
return ans
}
We can observe that if the first and last houses have different colors, the maximum distance is $n - 1$.

func abs(x int) int {
if x >= 0 {
return x
}
return -x
}
```
If the first and last houses have the same color, we can scan from the left to find the first house with a different color (let its index be $i$), and scan from the right to find the first house with a different color (let its index be $j$). The maximum distance is then $\max(n - i - 1, j)$.

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### Solution 2
The time complexity is $O(n)$, where $n$ is the number of houses. The space complexity is $O(1)$.

<!-- tabs:start -->

Expand Down Expand Up @@ -181,11 +107,13 @@ class Solution {
if (colors[0] != colors[n - 1]) {
return n - 1;
}
int i = 0, j = n - 1;
while (colors[++i] == colors[0])
;
while (colors[--j] == colors[0])
;
int i = 1, j = n - 2;
while (colors[i] == colors[0]) {
++i;
}
while (colors[j] == colors[0]) {
--j;
}
return Math.max(n - i - 1, j);
}
}
Expand All @@ -198,12 +126,16 @@ class Solution {
public:
int maxDistance(vector<int>& colors) {
int n = colors.size();
if (colors[0] != colors[n - 1]) return n - 1;
int i = 0, j = n;
while (colors[++i] == colors[0])
;
while (colors[--j] == colors[0])
;
if (colors[0] != colors[n - 1]) {
return n - 1;
}
int i = 1, j = n - 2;
while (colors[i] == colors[0]) {
++i;
}
while (colors[j] == colors[0]) {
--j;
}
return max(n - i - 1, j);
}
};
Expand All @@ -228,6 +160,47 @@ func maxDistance(colors []int) int {
}
```

#### TypeScript

```ts
function maxDistance(colors: number[]): number {
const n = colors.length;
if (colors[0] !== colors[n - 1]) {
return n - 1;
}
let [i, j] = [1, n - 2];
while (colors[i] === colors[0]) {
i++;
}
while (colors[j] === colors[0]) {
j--;
}
return Math.max(n - i - 1, j);
};
```

#### Rust

```rust
impl Solution {
pub fn max_distance(colors: Vec<i32>) -> i32 {
let n = colors.len();
if colors[0] != colors[n - 1] {
return (n - 1) as i32;
}
let mut i = 1;
while colors[i] == colors[0] {
i += 1;
}
let mut j = n - 2;
while colors[j] == colors[0] {
j -= 1;
}
std::cmp::max((n - i - 1) as i32, j as i32)
}
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
20 changes: 13 additions & 7 deletions solution/2000-2099/2078.Two Furthest Houses With Different Colors/Solution.cpp
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,11 +1,17 @@
class Solution {
public:
int maxDistance(vector<int>& colors) {
int ans = 0, n = colors.size();
for (int i = 0; i < n; ++i)
for (int j = i + 1; j < n; ++j)
if (colors[i] != colors[j])
ans = max(ans, abs(i - j));
return ans;
int n = colors.size();
if (colors[0] != colors[n - 1]) {
return n - 1;
}
int i = 1, j = n - 2;
while (colors[i] == colors[0]) {
++i;
}
while (colors[j] == colors[0]) {
--j;
}
return max(n - i - 1, j);
}
};
};
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