Merge branch 'youngyangyang04:master' into master

problems/0017.电话号码的字母组合.md

@@ -260,7 +260,7 @@ class Solution {
 
     }
 
-    //每次迭代获取一个字符串，所以会设计大量的字符串拼接，所以这里选择更为高效的 StringBuilder
+    //每次迭代获取一个字符串，所以会涉及大量的字符串拼接，所以这里选择更为高效的 StringBuilder
     StringBuilder temp = new StringBuilder();
 
     //比如digits如果为"23",num 为0，则str表示2对应的 abc
@@ -274,7 +274,7 @@ class Solution {
         String str = numString[digits.charAt(num) - '0'];
         for (int i = 0; i < str.length(); i++) {
             temp.append(str.charAt(i));
-            //c
+            //递归，处理下一层
             backTracking(digits, numString, num + 1);
             //剔除末尾的继续尝试
             temp.deleteCharAt(temp.length() - 1);

problems/0035.搜索插入位置.md

@@ -313,18 +313,18 @@ func searchInsert(nums []int, target int) int {
 
 ```rust
 impl Solution {
-    pub fn search_insert(nums: Vec<i32>, target: i32) -> i32 {
-        let mut left = 0;
-        let mut right = nums.len();
-        while left < right {
+      pub fn search_insert(nums: Vec<i32>, target: i32) -> i32 {
+        use std::cmp::Ordering::{Equal, Greater, Less};
+        let (mut left, mut right) = (0, nums.len() as i32 - 1);
+        while left <= right {
             let mid = (left + right) / 2;
-            match nums[mid].cmp(&target) {
-                Ordering::Less => left = mid + 1,
-                Ordering::Equal => return ((left + right) / 2) as i32,
-                Ordering::Greater => right = mid,
+            match nums[mid as usize].cmp(&target) {
+                Less => left = mid + 1,
+                Equal => return mid,
+                Greater => right = mid - 1,
             }
         }
-        ((left + right) / 2) as i32
+        right + 1
     }
 }
 ```

problems/0203.移除链表元素.md

@@ -585,7 +585,7 @@ impl Solution {
         let mut dummyHead = Box::new(ListNode::new(0));
         dummyHead.next = head;
         let mut cur = dummyHead.as_mut();
-	// 使用take()替换std::men::replace(&mut node.next, None)达到相同的效果，并且更普遍易读
+	// 使用take()替换std::mem::replace(&mut node.next, None)达到相同的效果，并且更普遍易读
         while let Some(nxt) = cur.next.take() {
             if nxt.val == val {
                 cur.next = nxt.next;