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Commit 29d3201

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feat: add solutions to lc problems: No.0670,3010 (doocs#2245)
* No.0670.Maximum Swap * No.3010.Divide an Array Into Subarrays With Minimum Cost I
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9 files changed

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‎solution/0600-0699/0670.Maximum Swap/README.md‎

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### 方法一:贪心
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先将数字转为字符串 `s`,然后从右往左遍历字符串 `s`,用数组或哈希表 `d` 记录每个数字右侧的最大数字的位置(可以是字符本身的位置)。
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我们先将数字转为字符串 $s$,然后从右往左遍历字符串 $s$,用数组或哈希表 $d$ 记录每个数字右侧的最大数字的位置(可以是数字本身的位置)。
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接着从左到右遍历 `d`,如果 `s[i] < s[d[i]]`,则进行交换,并退出遍历的过程。
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接着从左到右遍历 $d$,如果 $s[i] \lt s[d[i]]$,则进行交换,并退出遍历的过程。
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最后将字符串 `s` 转为数字,即为答案。
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最后将字符串 $s$ 转为数字,即为答案。
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时间复杂度 $O(\log C),ドル空间复杂度 $O(\log C)$。其中 $C$ 是数字 `num` 的值域。
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时间复杂度 $O(\log M),ドル空间复杂度 $O(\log M)$。其中 $M$ 是数字 $num$ 的取值范围。
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<!-- tabs:start -->
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‎solution/0600-0699/0670.Maximum Swap/README_EN.md‎

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## Solutions
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### Solution 1
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### Solution 1: Greedy Algorithm
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First, we convert the number into a string $s$. Then, we traverse the string $s$ from right to left, using an array or hash table $d$ to record the position of the maximum number to the right of each number (it can be the position of the number itself).
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Next, we traverse $d$ from left to right. If $s[i] < s[d[i]],ドル we swap them and exit the traversal process.
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Finally, we convert the string $s$ back into a number, which is the answer.
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The time complexity is $O(\log M),ドル and the space complexity is $O(\log M)$. Here, $M$ is the range of the number $num$.
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<!-- tabs:start -->
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‎solution/3000-3099/3010.Divide an Array Into Subarrays With Minimum Cost I/README.md‎

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## 解法
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### 方法一
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### 方法一:遍历找最小值和次小值
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我们记数组 $nums$ 的第一个元素为 $a,ドル其余元素中最小的元素为 $b,ドル次小的元素为 $c,ドル那么答案就是 $a+b+c$。
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时间复杂度 $O(n),ドル其中 $n$ 是数组 $nums$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```python
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class Solution:
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def minimumCost(self, nums: List[int]) -> int:
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a, b, c = nums[0], inf, inf
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for x in nums[1:]:
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if x < b:
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c, b = b, x
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elif x < c:
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c = x
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return a + b + c
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```
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```java
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class Solution {
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public int minimumCost(int[] nums) {
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int a = nums[0], b = 100, c = 100;
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for (int i = 1; i < nums.length; ++i) {
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if (nums[i] < b) {
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c = b;
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b = nums[i];
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} else if (nums[i] < c) {
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c = nums[i];
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}
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}
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return a + b + c;
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}
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}
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```
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```cpp
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class Solution {
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public:
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int minimumCost(vector<int>& nums) {
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int a = nums[0], b = 100, c = 100;
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for (int i = 1; i < nums.size(); ++i) {
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if (nums[i] < b) {
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c = b;
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b = nums[i];
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} else if (nums[i] < c) {
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c = nums[i];
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}
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}
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return a + b + c;
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}
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};
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```
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```go
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func minimumCost(nums []int) int {
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a, b, c := nums[0], 100, 100
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for _, x := range nums[1:] {
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if x < b {
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b, c = x, b
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} else if x < c {
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c = x
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}
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}
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return a + b + c
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}
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```
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```ts
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function minimumCost(nums: number[]): number {
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let [a, b, c] = [nums[0], 100, 100];
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for (const x of nums.slice(1)) {
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if (x < b) {
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[b, c] = [x, b];
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} else if (x < c) {
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c = x;
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}
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}
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return a + b + c;
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}
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```
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<!-- tabs:end -->

‎solution/3000-3099/3010.Divide an Array Into Subarrays With Minimum Cost I/README_EN.md‎

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## Solutions
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### Solution 1
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### Solution 1: Traverse to Find the Smallest and Second Smallest Values
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We set the first element of the array $nums$ as $a,ドル the smallest element among the remaining elements as $b,ドル and the second smallest element as $c$. The answer is $a+b+c$.
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The time complexity is $O(n),ドル where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
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<!-- tabs:start -->
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```python
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class Solution:
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def minimumCost(self, nums: List[int]) -> int:
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a, b, c = nums[0], inf, inf
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for x in nums[1:]:
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if x < b:
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c, b = b, x
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elif x < c:
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c = x
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return a + b + c
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```
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```java
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class Solution {
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public int minimumCost(int[] nums) {
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int a = nums[0], b = 100, c = 100;
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for (int i = 1; i < nums.length; ++i) {
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if (nums[i] < b) {
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c = b;
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b = nums[i];
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} else if (nums[i] < c) {
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c = nums[i];
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}
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}
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return a + b + c;
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}
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}
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```
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```cpp
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class Solution {
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public:
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int minimumCost(vector<int>& nums) {
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int a = nums[0], b = 100, c = 100;
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for (int i = 1; i < nums.size(); ++i) {
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if (nums[i] < b) {
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c = b;
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b = nums[i];
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} else if (nums[i] < c) {
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c = nums[i];
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}
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}
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return a + b + c;
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}
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};
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```
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```go
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func minimumCost(nums []int) int {
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a, b, c := nums[0], 100, 100
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for _, x := range nums[1:] {
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if x < b {
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b, c = x, b
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} else if x < c {
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c = x
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}
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}
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return a + b + c
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}
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```
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```ts
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function minimumCost(nums: number[]): number {
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let [a, b, c] = [nums[0], 100, 100];
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for (const x of nums.slice(1)) {
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if (x < b) {
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[b, c] = [x, b];
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} else if (x < c) {
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c = x;
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}
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}
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return a + b + c;
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}
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```
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<!-- tabs:end -->
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class Solution {
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public:
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int minimumCost(vector<int>& nums) {
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int a = nums[0], b = 100, c = 100;
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for (int i = 1; i < nums.size(); ++i) {
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if (nums[i] < b) {
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c = b;
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b = nums[i];
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} else if (nums[i] < c) {
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c = nums[i];
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}
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}
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return a + b + c;
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}
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};
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func minimumCost(nums []int) int {
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a, b, c := nums[0], 100, 100
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for _, x := range nums[1:] {
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if x < b {
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b, c = x, b
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} else if x < c {
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c = x
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}
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}
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return a + b + c
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}
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class Solution {
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public int minimumCost(int[] nums) {
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int a = nums[0], b = 100, c = 100;
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for (int i = 1; i < nums.length; ++i) {
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if (nums[i] < b) {
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c = b;
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b = nums[i];
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} else if (nums[i] < c) {
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c = nums[i];
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}
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}
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return a + b + c;
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}
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}
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class Solution:
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def minimumCost(self, nums: List[int]) -> int:
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a, b, c = nums[0], inf, inf
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for x in nums[1:]:
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if x < b:
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c, b = b, x
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elif x < c:
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c = x
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return a + b + c
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function minimumCost(nums: number[]): number {
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let [a, b, c] = [nums[0], 100, 100];
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for (const x of nums.slice(1)) {
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if (x < b) {
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[b, c] = [x, b];
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} else if (x < c) {
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c = x;
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}
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}
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return a + b + c;
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}

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