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Commit c6953e5

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feat: update solutions to lc problems: No.1171,1891 (#1843)
* No.1171.Remove Zero Sum Consecutive Nodes from Linked List * No.1891.Cutting Ribbons
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‎solution/1100-1199/1171.Remove Zero Sum Consecutive Nodes from Linked List/README.md‎

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### **Rust**
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```
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```rust
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// Definition for singly-linked list.
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// #[derive(PartialEq, Eq, Clone, Debug)]
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// pub struct ListNode {

‎solution/1100-1199/1171.Remove Zero Sum Consecutive Nodes from Linked List/README_EN.md‎

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## Solutions
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**Solution 1: Prefix Sum + Hash Table**
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If two prefix sums of the linked list are equal, it means that the sum of the continuous node sequence between the two prefix sums is 0ドル,ドル so we can remove this part of the continuous nodes.
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We first traverse the linked list and use a hash table $last$ to record the prefix sum and the corresponding linked list node. For the same prefix sum $s,ドル the later node overwrites the previous node.
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Next, we traverse the linked list again. If the current node $cur$ has a prefix sum $s$ that appears in $last,ドル it means that the sum of all nodes between $cur$ and $last[s]$ is 0ドル,ドル so we directly modify the pointer of $cur$ to $last[s].next,ドル which removes this part of the continuous nodes with a sum of 0ドル$. We continue to traverse and delete all continuous nodes with a sum of 0ドル$.
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Finally, we return the head node of the linked list $dummy.next$.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the length of the linked list.
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### **Python3**
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### **Rust**
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```
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```rust
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// Definition for singly-linked list.
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// #[derive(PartialEq, Eq, Clone, Debug)]
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// pub struct ListNode {

‎solution/1800-1899/1891.Cutting Ribbons/README.md‎

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最后,我们返回 $left$ 即可。
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时间复杂度 $O(n \times \log M),ドル空间复杂度 $O(1)$。其中 $n$ 和 $M$ 分别为绳子的数量和绳子的最大长度。
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时间复杂度 $O(n \times \log M),ドル其中 $n$ 和 $M$ 分别为绳子的数量和绳子的最大长度。空间复杂度 $O(1)$。
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```rust
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impl Solution {
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fn max_length(ribbons: Vec<i32>, k: i32) -> i32 {
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pubfn max_length(ribbons: Vec<i32>, k: i32) -> i32 {
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let mut left = 0i32;
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let mut right = *ribbons.iter().max().unwrap();
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while left < right {

‎solution/1800-1899/1891.Cutting Ribbons/README_EN.md‎

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## Solutions
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**Solution 1: Binary Search**
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We observe that if we can obtain $k$ ropes of length $x,ドル then we can also obtain $k$ ropes of length $x-1$. This implies that there is a monotonicity property, and we can use binary search to find the maximum length $x$ such that we can obtain $k$ ropes of length $x$.
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We define the left boundary of the binary search as $left=0,ドル the right boundary as $right=\max(ribbons),ドル and the middle value as $mid=(left+right+1)/2$. We then calculate the number of ropes we can obtain with length $mid,ドル denoted as $cnt$. If $cnt \geq k,ドル it means we can obtain $k$ ropes of length $mid,ドル so we update $left$ to $mid$. Otherwise, we update $right$ to $mid-1$.
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Finally, we return $left$ as the maximum length of the ropes we can obtain.
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The time complexity is $O(n \times \log M),ドル where $n$ and $M$ are the number of ropes and the maximum length of the ropes, respectively. The space complexity is $O(1)$.
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### **Python3**
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```rust
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impl Solution {
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fn max_length(ribbons: Vec<i32>, k: i32) -> i32 {
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pubfn max_length(ribbons: Vec<i32>, k: i32) -> i32 {
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let mut left = 0i32;
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let mut right = *ribbons.iter().max().unwrap();
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while left < right {

‎solution/1800-1899/1891.Cutting Ribbons/Solution.rs‎

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impl Solution {
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fn max_length(ribbons: Vec<i32>, k: i32) -> i32 {
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pubfn max_length(ribbons: Vec<i32>, k: i32) -> i32 {
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let mut left = 0i32;
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let mut right = *ribbons.iter().max().unwrap();
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while left < right {

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