Merge branch 'youngyangyang04:master' into master

problems/0131.分割回文串.md

@@ -587,5 +587,57 @@ func partition(_ s: String) -> [[String]] {
 }
 ```
 
+## Rust
+
+```rust
+impl Solution {
+    pub fn partition(s: String) -> Vec<Vec<String>> {
+        let mut ret = vec![];
+        let mut path = vec![];
+        let sub_str: Vec<char> = s.chars().collect();
+        
+        Self::backtracing(&sub_str, 0, &mut ret, &mut path);
+        
+        ret
+    }
+    
+    fn backtracing(sub_str: &Vec<char>, start: usize, ret: &mut Vec<Vec<String>>, path: &mut Vec<String>) {
+        //如果起始位置大于s的大小，说明找到了一组分割方案
+        if start >= sub_str.len() {
+            ret.push(path.clone());
+            return;
+        }
+        
+        for i in start..sub_str.len() {
+            if !Self::is_palindrome(sub_str, start, i) {
+                continue;
+            }
+            //如果是回文子串，则记录
+            let s: String = sub_str[start..i+1].into_iter().collect();
+            path.push(s);
+            
+            //起始位置后移，保证不重复
+            Self::backtracing(sub_str, i+1, ret, path); 
+            path.pop();
+        }
+        
+    }
+    
+    fn is_palindrome(s: &Vec<char>, start: usize, end: usize) -> bool {
+        let (mut start, mut end) = (start, end);
+        
+        while start < end {
+            if s[start] != s[end] {
+                return false;
+            }
+            
+            start += 1;
+            end -= 1;
+        }
+
+        true
+    }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0150.逆波兰表达式求值.md

@@ -133,39 +133,26 @@ public:
 java:
 
 ```Java
-public class EvalRPN {
-
+class Solution {
     public int evalRPN(String[] tokens) {
         Deque<Integer> stack = new LinkedList();
-        for (String token : tokens) {
-            char c = token.charAt(0);
-            if (!isOpe(token)) {
-                stack.addFirst(stoi(token));
-            } else if (c == '+') {
-                stack.push(stack.pop() + stack.pop());
-            } else if (c == '-') {
-                stack.push(- stack.pop() + stack.pop());
-            } else if (c == '*') {
-                stack.push( stack.pop() * stack.pop());
+        for (int i = 0; i < tokens.length; ++i) {
+            if ("+".equals(tokens[i])) {        // leetcode 内置jdk的问题，不能使用==判断字符串是否相等
+                stack.push(stack.pop() + stack.pop());      // 注意 - 和/ 需要特殊处理
+            } else if ("-".equals(tokens[i])) {
+                stack.push(-stack.pop() + stack.pop());
+            } else if ("*".equals(tokens[i])) {
+                stack.push(stack.pop() * stack.pop());
+            } else if ("/".equals(tokens[i])) {
+                int temp1 = stack.pop();
+                int temp2 = stack.pop();
+                stack.push(temp2 / temp1);
             } else {
-                int num1 = stack.pop();
-                int num2 = stack.pop();
-                stack.push( num2/num1);
+                stack.push(Integer.valueOf(tokens[i]));
             }
         }
         return stack.pop();
     }
-    private boolean isOpe(String s) {
-        return s.length() == 1 && s.charAt(0) <'0' || s.charAt(0) >'9';
-    }
-    private int stoi(String s) {
-        return Integer.valueOf(s);
-    }
-
-    public static void main(String[] args) {
-        new EvalRPN().evalRPN(new String[] {"10","6","9","3","+","-11","*","/","*","17","+","5","+"});
-    }
-
 }
 ```
 

problems/0459.重复的子字符串.md

@@ -45,7 +45,7 @@
 
 这里就要说一说next数组了，next 数组记录的就是最长相同前后缀( [字符串：KMP算法精讲](https://programmercarl.com/0028.实现strStr.html) 这里介绍了什么是前缀，什么是后缀，什么又是最长相同前后缀)， 如果 next[len - 1] != -1，则说明字符串有最长相同的前后缀（就是字符串里的前缀子串和后缀子串相同的最长长度）。
 
-最长相等前后缀的长度为：next[len - 1] + 1。
+最长相等前后缀的长度为：next[len - 1] + 1。(这里的next数组是以统一减一的方式计算的，因此需要+1)
 
 数组长度为：len。