Merge branch 'master' of github.com:youngyangyang04/leetcode-master

problems/0090.子集II.md

@@ -440,7 +440,7 @@ impl Solution {
         let len = nums.len();
         // if start_index >= len { return; }
         for i in start_index..len {
-            if i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false { continue; }
+            if i > 0 && nums[i] == nums[i - 1] && !used[i - 1] { continue; }
             path.push(nums[i]);
             used[i] = true;
             Self::backtracking(result, path, nums, i + 1, used);
@@ -449,11 +449,10 @@ impl Solution {
         }
     }
 
-    pub fn subsets_with_dup(nums: Vec<i32>) -> Vec<Vec<i32>> {
+    pub fn subsets_with_dup(mut nums: Vec<i32>) -> Vec<Vec<i32>> {
         let mut result: Vec<Vec<i32>> = Vec::new();
         let mut path: Vec<i32> = Vec::new();
         let mut used = vec![false; nums.len()];
-        let mut nums = nums;
         nums.sort();
         Self::backtracking(&mut result, &mut path, &nums, 0, &mut used);
         result
@@ -461,6 +460,35 @@ impl Solution {
 }
 ```
 
+set 去重版本
+
+```rust
+use std::collections::HashSet;
+impl Solution {
+    pub fn subsets_with_dup(mut nums: Vec<i32>) -> Vec<Vec<i32>> {
+        let mut res = HashSet::new();
+        let mut path = vec![];
+        nums.sort();
+        Self::backtracking(&nums, &mut path, &mut res, 0);
+        res.into_iter().collect()
+    }
+
+    pub fn backtracking(
+        nums: &Vec<i32>,
+        path: &mut Vec<i32>,
+        res: &mut HashSet<Vec<i32>>,
+        start_index: usize,
+    ) {
+        res.insert(path.clone());
+        for i in start_index..nums.len() {
+            path.push(nums[i]);
+            Self::backtracking(nums, path, res, i + 1);
+            path.pop();
+        }
+    }
+}
+```
+
 ### C 
 
 ```c

problems/0203.移除链表元素.md

@@ -5,7 +5,7 @@
 <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>
 
 
-> 链表操作中，可以使用原链表来直接进行删除操作，也可以设置一个虚拟头结点在进行删除操作，接下来看一看哪种方式更方便。
+> 链表操作中，可以使用原链表来直接进行删除操作，也可以设置一个虚拟头结点再进行删除操作，接下来看一看哪种方式更方便。
 
 # 203.移除链表元素
 

problems/1207.独一无二的出现次数.md

@@ -99,6 +99,7 @@ class Solution {
 
 Python：
 ```python 
+# 方法 1: 数组在哈西法的应用
 class Solution:
     def uniqueOccurrences(self, arr: List[int]) -> bool:
         count = [0] * 2002
@@ -113,6 +114,26 @@ class Solution:
                     return False
         return True
 ```
+```python 
+# 方法 2： map 在哈西法的应用
+class Solution:
+    def uniqueOccurrences(self, arr: List[int]) -> bool:
+        ref = dict()
+
+        for i in range(len(arr)):
+            ref[arr[i]] = ref.get(arr[i], 0) + 1
+
+        value_list = sorted(ref.values())
+
+        for i in range(len(value_list) - 1):
+            if value_list[i + 1] == value_list[i]:
+                return False 
+
+        return True 
+
+```
+
+
 Go：
 
 JavaScript：

problems/二叉树理论基础.md

@@ -59,6 +59,7 @@
 * 它的左、右子树也分别为二叉排序树
 
 下面这两棵树都是搜索树
+
 <img src='https://img-blog.csdnimg.cn/20200806190304693.png' width=600> </img></div>
 
 

problems/算法模板.md

@@ -803,6 +803,46 @@ Java：
 
 
 Python：
+## 二分查找法 
+```python 
+def binarysearch(nums, target):
+    low = 0
+    high = len(nums) - 1
+    while (low <= high):
+        mid = (high + low)//2
+
+        if (nums[mid] < target):
+            low = mid + 1
+            
+        if (nums[mid] > target):
+            high = mid - 1
+            
+        if (nums[mid] == target):
+            return mid
+
+    return -1
+```
+
+## KMP
+
+```python 
+def kmp(self, a, s):
+    # a: length of the array
+    # s: string 
+    
+    next = [0]*a
+    j = 0    
+    next[0] = 0
+ 
+    for i in range(1, len(s)):
+        while j > 0 and s[j] != s[i]:
+            j = next[j - 1]
+            
+        if s[j] == s[i]:
+            j += 1
+        next[i] = j
+    return next 
+```
 
 
 Go：