mirror of
https://github.com/youngyangyang04/leetcode-master.git
synced 2025-07-08 00:43:04 +08:00
Merge branch 'youngyangyang04:master' into master
This commit is contained in:
@ -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);
|
||||
|
@ -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
|
||||
}
|
||||
}
|
||||
```
|
||||
|
@ -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;
|
||||
|
Reference in New Issue
Block a user