mirror of
https://github.com/youngyangyang04/leetcode-master.git
synced 2025-07-09 19:44:45 +08:00
update: 0435.无重叠区间,rust版本代码报错
This commit is contained in:
@ -4,7 +4,6 @@
|
||||
</a>
|
||||
<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!</strong></p>
|
||||
|
||||
|
||||
# 435. 无重叠区间
|
||||
|
||||
[力扣题目链接](https://leetcode.cn/problems/non-overlapping-intervals/)
|
||||
@ -16,19 +15,22 @@
|
||||
区间 [1,2] 和 [2,3] 的边界相互“接触”,但没有相互重叠。
|
||||
|
||||
示例 1:
|
||||
* 输入: [ [1,2], [2,3], [3,4], [1,3] ]
|
||||
* 输出: 1
|
||||
* 解释: 移除 [1,3] 后,剩下的区间没有重叠。
|
||||
|
||||
- 输入: [ [1,2], [2,3], [3,4], [1,3] ]
|
||||
- 输出: 1
|
||||
- 解释: 移除 [1,3] 后,剩下的区间没有重叠。
|
||||
|
||||
示例 2:
|
||||
* 输入: [ [1,2], [1,2], [1,2] ]
|
||||
* 输出: 2
|
||||
* 解释: 你需要移除两个 [1,2] 来使剩下的区间没有重叠。
|
||||
|
||||
- 输入: [ [1,2], [1,2], [1,2] ]
|
||||
- 输出: 2
|
||||
- 解释: 你需要移除两个 [1,2] 来使剩下的区间没有重叠。
|
||||
|
||||
示例 3:
|
||||
* 输入: [ [1,2], [2,3] ]
|
||||
* 输出: 0
|
||||
* 解释: 你不需要移除任何区间,因为它们已经是无重叠的了。
|
||||
|
||||
- 输入: [ [1,2], [2,3] ]
|
||||
- 输出: 0
|
||||
- 解释: 你不需要移除任何区间,因为它们已经是无重叠的了。
|
||||
|
||||
## 算法公开课
|
||||
|
||||
@ -84,8 +86,9 @@ public:
|
||||
}
|
||||
};
|
||||
```
|
||||
* 时间复杂度:O(nlog n) ,有一个快排
|
||||
* 空间复杂度:O(n),有一个快排,最差情况(倒序)时,需要n次递归调用。因此确实需要O(n)的栈空间
|
||||
|
||||
- 时间复杂度:O(nlog n) ,有一个快排
|
||||
- 空间复杂度:O(n),有一个快排,最差情况(倒序)时,需要 n 次递归调用。因此确实需要 O(n)的栈空间
|
||||
|
||||
大家此时会发现如此复杂的一个问题,代码实现却这么简单!
|
||||
|
||||
@ -176,6 +179,7 @@ public:
|
||||
```
|
||||
|
||||
这里按照 左边界排序,或者按照右边界排序,都可以 AC,原理是一样的。
|
||||
|
||||
```CPP
|
||||
class Solution {
|
||||
public:
|
||||
@ -204,8 +208,8 @@ public:
|
||||
|
||||
## 其他语言版本
|
||||
|
||||
|
||||
### Java
|
||||
|
||||
```java
|
||||
class Solution {
|
||||
public int eraseOverlapIntervals(int[][] intervals) {
|
||||
@ -227,6 +231,7 @@ class Solution {
|
||||
```
|
||||
|
||||
按左边排序,不管右边顺序。相交的时候取最小的右边。
|
||||
|
||||
```java
|
||||
class Solution {
|
||||
public int eraseOverlapIntervals(int[][] intervals) {
|
||||
@ -248,7 +253,9 @@ class Solution {
|
||||
```
|
||||
|
||||
### Python
|
||||
|
||||
贪心 基于左边界
|
||||
|
||||
```python
|
||||
class Solution:
|
||||
def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
|
||||
@ -266,7 +273,9 @@ class Solution:
|
||||
return count
|
||||
|
||||
```
|
||||
|
||||
贪心 基于左边界 把 452.用最少数量的箭引爆气球代码稍做修改
|
||||
|
||||
```python
|
||||
class Solution:
|
||||
def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
|
||||
@ -287,7 +296,9 @@ class Solution:
|
||||
|
||||
|
||||
```
|
||||
|
||||
### Go
|
||||
|
||||
```go
|
||||
func eraseOverlapIntervals(intervals [][]int) int {
|
||||
sort.Slice(intervals, func(i, j int) bool {
|
||||
@ -312,7 +323,9 @@ func min(a, b int) int {
|
||||
```
|
||||
|
||||
### Javascript
|
||||
|
||||
- 按右边界排序
|
||||
|
||||
```Javascript
|
||||
var eraseOverlapIntervals = function(intervals) {
|
||||
intervals.sort((a, b) => {
|
||||
@ -333,23 +346,25 @@ var eraseOverlapIntervals = function(intervals) {
|
||||
return intervals.length - count
|
||||
};
|
||||
```
|
||||
|
||||
- 按左边界排序
|
||||
|
||||
```js
|
||||
var eraseOverlapIntervals = function (intervals) {
|
||||
// 按照左边界升序排列
|
||||
intervals.sort((a, b) => a[0] - b[0])
|
||||
let count = 1
|
||||
let end = intervals[intervals.length - 1][0]
|
||||
intervals.sort((a, b) => a[0] - b[0]);
|
||||
let count = 1;
|
||||
let end = intervals[intervals.length - 1][0];
|
||||
// 倒序遍历,对单个区间来说,左边界越大越好,因为给前面区间的空间越大
|
||||
for (let i = intervals.length - 2; i >= 0; i--) {
|
||||
if (intervals[i][1] <= end) {
|
||||
count++
|
||||
end = intervals[i][0]
|
||||
count++;
|
||||
end = intervals[i][0];
|
||||
}
|
||||
}
|
||||
// count 记录的是最大非重复区间的个数
|
||||
return intervals.length - count
|
||||
}
|
||||
return intervals.length - count;
|
||||
};
|
||||
```
|
||||
|
||||
### TypeScript
|
||||
@ -370,7 +385,7 @@ function eraseOverlapIntervals(intervals: number[][]): number {
|
||||
}
|
||||
}
|
||||
return length - count;
|
||||
};
|
||||
}
|
||||
```
|
||||
|
||||
> 按左边界排序,从左往右遍历
|
||||
@ -394,7 +409,7 @@ function eraseOverlapIntervals(intervals: number[][]): number {
|
||||
}
|
||||
}
|
||||
return resCount;
|
||||
};
|
||||
}
|
||||
```
|
||||
|
||||
### Scala
|
||||
@ -423,7 +438,7 @@ object Solution {
|
||||
|
||||
```Rust
|
||||
impl Solution {
|
||||
pub fn erase_overlap_intervals(intervals: Vec<Vec<i32>>) -> i32 {
|
||||
pub fn erase_overlap_intervals(mut intervals: Vec<Vec<i32>>) -> i32 {
|
||||
if intervals.is_empty() {
|
||||
return 0;
|
||||
}
|
||||
@ -441,7 +456,9 @@ impl Solution {
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
### C#
|
||||
|
||||
```csharp
|
||||
public class Solution
|
||||
{
|
||||
@ -463,9 +480,7 @@ public class Solution
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
<p align="center">
|
||||
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
|
||||
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
|
||||
</a>
|
||||
|
||||
|
Reference in New Issue
Block a user