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feat: add solutions to lc problems: No.2855~2858 #1639

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# [2855. 使数组成为递增数组的最少右移次数](https://leetcode.cn/problems/minimum-right-shifts-to-sort-the-array)

[English Version](/solution/2800-2899/2855.Minimum%20Right%20Shifts%20to%20Sort%20the%20Array/README_EN.md)

## 题目描述

<!-- 这里写题目描述 -->

<p>给你一个长度为 <code>n</code>&nbsp;下标从 <strong>0</strong>&nbsp;开始的数组&nbsp;<code>nums</code>&nbsp;,数组中的元素为&nbsp;<strong>互不相同</strong>&nbsp;的正整数。请你返回让 <code>nums</code>&nbsp;成为递增数组的 <strong>最少右移</strong>&nbsp;次数,如果无法得到递增数组,返回 <code>-1</code>&nbsp;。</p>

<p>一次 <strong>右移</strong>&nbsp;指的是同时对所有下标进行操作,将下标为 <code>i</code>&nbsp;的元素移动到下标&nbsp;<code>(i + 1) % n</code>&nbsp;处。</p>

<p>&nbsp;</p>

<p><strong class="example">示例 1:</strong></p>

<pre>
<b>输入:</b>nums = [3,4,5,1,2]
<b>输出:</b>2
<b>解释:</b>
第一次右移后,nums = [2,3,4,5,1] 。
第二次右移后,nums = [1,2,3,4,5] 。
现在 nums 是递增数组了,所以答案为 2 。
</pre>

<p><strong class="example">示例 2:</strong></p>

<pre>
<b>输入:</b>nums = [1,3,5]
<b>输出:</b>0
<b>解释:</b>nums 已经是递增数组了,所以答案为 0 。</pre>

<p><strong class="example">示例 3:</strong></p>

<pre>
<b>输入:</b>nums = [2,1,4]
<b>输出:</b>-1
<b>解释:</b>无法将数组变为递增数组。
</pre>

<p>&nbsp;</p>

<p><strong>提示:</strong></p>

<ul>
<li><code>1 &lt;= nums.length &lt;= 100</code></li>
<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
<li><code>nums</code>&nbsp;中的整数互不相同。</li>
</ul>

## 解法

<!-- 这里可写通用的实现逻辑 -->

**方法一:直接遍历**

我们先用一个指针 $i$ 从左到右遍历数组 $nums,ドル找出一段连续的递增序列,直到 $i$ 到达数组末尾或者 $nums[i - 1] \gt nums[i]$。接下来我们用另一个指针 $k$ 从 $i + 1$ 开始遍历数组 $nums,ドル找出一段连续的递增序列,直到 $k$ 到达数组末尾或者 $nums[k - 1] \gt nums[k]$ 且 $nums[k] \gt nums[0]$。如果 $k$ 到达数组末尾,说明数组已经是递增的,返回 $n - i$;否则返回 $-1$。

时间复杂度 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 是数组 $nums$ 的长度。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python
class Solution:
def minimumRightShifts(self, nums: List[int]) -> int:
n = len(nums)
i = 1
while i < n and nums[i - 1] < nums[i]:
i += 1
k = i + 1
while k < n and nums[k - 1] < nums[k] < nums[0]:
k += 1
return -1 if k < n else n - i
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java
class Solution {
public int minimumRightShifts(List<Integer> nums) {
int n = nums.size();
int i = 1;
while (i < n && nums.get(i - 1) < nums.get(i)) {
++i;
}
int k = i + 1;
while (k < n && nums.get(k - 1) < nums.get(k) && nums.get(k) < nums.get(0)) {
++k;
}
return k < n ? -1 : n - i;
}
}
```

### **C++**

```cpp
class Solution {
public:
int minimumRightShifts(vector<int>& nums) {
int n = nums.size();
int i = 1;
while (i < n && nums[i - 1] < nums[i]) {
++i;
}
int k = i + 1;
while (k < n && nums[k - 1] < nums[k] && nums[k] < nums[0]) {
++k;
}
return k < n ? -1 : n - i;
}
};
```

### **Go**

```go
func minimumRightShifts(nums []int) int {
n := len(nums)
i := 1
for i < n && nums[i-1] < nums[i] {
i++
}
k := i + 1
for k < n && nums[k-1] < nums[k] && nums[k] < nums[0] {
k++
}
if k < n {
return -1
}
return n - i
}
```

### **TypeScript**

```ts
function minimumRightShifts(nums: number[]): number {
const n = nums.length;
let i = 1;
while (i < n && nums[i - 1] < nums[i]) {
++i;
}
let k = i + 1;
while (k < n && nums[k - 1] < nums[k] && nums[k] < nums[0]) {
++k;
}
return k < n ? -1 : n - i;
}
```

### **...**

```

```

<!-- tabs:end -->
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# [2855. Minimum Right Shifts to Sort the Array](https://leetcode.com/problems/minimum-right-shifts-to-sort-the-array)

[中文文档](/solution/2800-2899/2855.Minimum%20Right%20Shifts%20to%20Sort%20the%20Array/README.md)

## Description

<p>You are given a <strong>0-indexed</strong> array <code>nums</code> of length <code>n</code> containing <strong>distinct</strong> positive integers. Return <em>the <strong>minimum</strong> number of <strong>right shifts</strong> required to sort </em><code>nums</code><em> and </em><code>-1</code><em> if this is not possible.</em></p>

<p>A <strong>right shift</strong> is defined as shifting the element at index <code>i</code> to index <code>(i + 1) % n</code>, for all indices.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,4,5,1,2]
<strong>Output:</strong> 2
<strong>Explanation:</strong>
After the first right shift, nums = [2,3,4,5,1].
After the second right shift, nums = [1,2,3,4,5].
Now nums is sorted; therefore the answer is 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,5]
<strong>Output:</strong> 0
<strong>Explanation:</strong> nums is already sorted therefore, the answer is 0.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,1,4]
<strong>Output:</strong> -1
<strong>Explanation:</strong> It&#39;s impossible to sort the array using right shifts.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
<li><code>1 &lt;= nums.length &lt;= 100</code></li>
<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
<li><code>nums</code> contains distinct integers.</li>
</ul>

## Solutions

<!-- tabs:start -->

### **Python3**

```python
class Solution:
def minimumRightShifts(self, nums: List[int]) -> int:
n = len(nums)
i = 1
while i < n and nums[i - 1] < nums[i]:
i += 1
k = i + 1
while k < n and nums[k - 1] < nums[k] < nums[0]:
k += 1
return -1 if k < n else n - i
```

### **Java**

```java
class Solution {
public int minimumRightShifts(List<Integer> nums) {
int n = nums.size();
int i = 1;
while (i < n && nums.get(i - 1) < nums.get(i)) {
++i;
}
int k = i + 1;
while (k < n && nums.get(k - 1) < nums.get(k) && nums.get(k) < nums.get(0)) {
++k;
}
return k < n ? -1 : n - i;
}
}
```

### **C++**

```cpp
class Solution {
public:
int minimumRightShifts(vector<int>& nums) {
int n = nums.size();
int i = 1;
while (i < n && nums[i - 1] < nums[i]) {
++i;
}
int k = i + 1;
while (k < n && nums[k - 1] < nums[k] && nums[k] < nums[0]) {
++k;
}
return k < n ? -1 : n - i;
}
};
```

### **Go**

```go
func minimumRightShifts(nums []int) int {
n := len(nums)
i := 1
for i < n && nums[i-1] < nums[i] {
i++
}
k := i + 1
for k < n && nums[k-1] < nums[k] && nums[k] < nums[0] {
k++
}
if k < n {
return -1
}
return n - i
}
```

### **TypeScript**

```ts
function minimumRightShifts(nums: number[]): number {
const n = nums.length;
let i = 1;
while (i < n && nums[i - 1] < nums[i]) {
++i;
}
let k = i + 1;
while (k < n && nums[k - 1] < nums[k] && nums[k] < nums[0]) {
++k;
}
return k < n ? -1 : n - i;
}
```

### **...**

```

```

<!-- tabs:end -->
View file Open in desktop
Original file line number Diff line number Diff line change
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class Solution {
public:
int minimumRightShifts(vector<int>& nums) {
int n = nums.size();
int i = 1;
while (i < n && nums[i - 1] < nums[i]) {
++i;
}
int k = i + 1;
while (k < n && nums[k - 1] < nums[k] && nums[k] < nums[0]) {
++k;
}
return k < n ? -1 : n - i;
}
};
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,15 @@
func minimumRightShifts(nums []int) int {
n := len(nums)
i := 1
for i < n && nums[i-1] < nums[i] {
i++
}
k := i + 1
for k < n && nums[k-1] < nums[k] && nums[k] < nums[0] {
k++
}
if k < n {
return -1
}
return n - i
}
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Original file line number Diff line number Diff line change
@@ -0,0 +1,14 @@
class Solution {
public int minimumRightShifts(List<Integer> nums) {
int n = nums.size();
int i = 1;
while (i < n && nums.get(i - 1) < nums.get(i)) {
++i;
}
int k = i + 1;
while (k < n && nums.get(k - 1) < nums.get(k) && nums.get(k) < nums.get(0)) {
++k;
}
return k < n ? -1 : n - i;
}
}
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,10 @@
class Solution:
def minimumRightShifts(self, nums: List[int]) -> int:
n = len(nums)
i = 1
while i < n and nums[i - 1] < nums[i]:
i += 1
k = i + 1
while k < n and nums[k - 1] < nums[k] < nums[0]:
k += 1
return -1 if k < n else n - i
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,12 @@
function minimumRightShifts(nums: number[]): number {
const n = nums.length;
let i = 1;
while (i < n && nums[i - 1] < nums[i]) {
++i;
}
let k = i + 1;
while (k < n && nums[k - 1] < nums[k] && nums[k] < nums[0]) {
++k;
}
return k < n ? -1 : n - i;
}
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