Merge branch 'youngyangyang04:master' into master

problems/0047.全排列II.md

@@ -323,5 +323,76 @@ func permuteUnique(_ nums: [Int]) -> [[Int]] {
 }
 ```
 
+### C
+```c
+//临时数组
+int *path;
+//返回数组
+int **ans;
+int *used;
+int pathTop, ansTop;
+
+//拷贝path到ans中
+void copyPath() {
+    int *tempPath = (int*)malloc(sizeof(int) * pathTop);
+    int i;
+    for(i = 0; i < pathTop; ++i) {
+        tempPath[i] = path[i];
+    }
+    ans[ansTop++] = tempPath;
+}
+
+void backTracking(int* used, int *nums, int numsSize) {
+    //若path中元素个数等于numsSize，将path拷贝入ans数组中
+    if(pathTop == numsSize)
+        copyPath();
+    int i;
+    for(i = 0; i < numsSize; i++) {
+        //若当前元素已被使用
+        //或前一位元素与当前元素值相同但并未被使用
+        //则跳过此分支
+        if(used[i] || (i != 0 && nums[i] == nums[i-1] && used[i-1] == 0))
+            continue;
+
+        //将当前元素的使用情况设为True
+        used[i] = 1;
+        path[pathTop++] = nums[i];
+        backTracking(used, nums, numsSize);
+        used[i] = 0;
+        --pathTop;
+    }
+}
+
+int cmp(void* elem1, void* elem2) {
+    return *((int*)elem1) - *((int*)elem2);
+}
+
+int** permuteUnique(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
+    //排序数组
+    qsort(nums, numsSize, sizeof(int), cmp);
+    //初始化辅助变量
+    pathTop = ansTop = 0;
+    path = (int*)malloc(sizeof(int) * numsSize);
+    ans = (int**)malloc(sizeof(int*) * 1000);
+    //初始化used辅助数组
+    used = (int*)malloc(sizeof(int) * numsSize);
+    int i;
+    for(i = 0; i < numsSize; i++) {
+        used[i] = 0;
+    }
+
+    backTracking(used, nums, numsSize);
+
+    //设置返回的数组的长度
+    *returnSize = ansTop;
+    *returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
+    int z;
+    for(z = 0; z < ansTop; z++) {
+        (*returnColumnSizes)[z] = numsSize;
+    }
+    return ans;
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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。