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对0034.在排序数组中查找元素的第一个和最后一个位置中python代码块语法进行修正,并添加新解法。补充0203.移除链表元素使用原先链表的python解法 #2901

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View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -397,38 +397,74 @@ class Solution:
def searchRange(self, nums: List[int], target: int) -> List[int]:
def getRightBorder(nums:List[int], target:int) -> int:
left, right = 0, len(nums)-1
rightBoder = -2 # 记录一下rightBorder没有被赋值的情况
rightBorder = -2 # 记录一下rightBorder没有被赋值的情况
while left <= right:
middle = left + (right-left) // 2
if nums[middle] > target:
right = middle - 1
else: # 寻找右边界,nums[middle] == target的时候更新left
left = middle + 1
rightBoder = left
rightBorder = left

return rightBoder
return rightBorder

def getLeftBorder(nums:List[int], target:int) -> int:
left, right = 0, len(nums)-1
leftBoder = -2 # 记录一下leftBorder没有被赋值的情况
leftBorder = -2 # 记录一下leftBorder没有被赋值的情况
while left <= right:
middle = left + (right-left) // 2
if nums[middle] >= target: # 寻找左边界,nums[middle] == target的时候更新right
right = middle - 1
leftBoder = right
leftBorder = right
else:
left = middle + 1
return leftBoder
leftBoder = getLeftBorder(nums, target)
rightBoder = getRightBorder(nums, target)
return leftBorder
leftBorder = getLeftBorder(nums, target)
rightBorder = getRightBorder(nums, target)
# 情况一
if leftBoder == -2 or rightBoder == -2: return [-1, -1]
if leftBorder == -2 or rightBorder == -2: return [-1, -1]
# 情况三
if rightBoder -leftBoder >1: return [leftBoder + 1, rightBoder - 1]
if rightBorder -leftBorder >1: return [leftBorder + 1, rightBorder - 1]
# 情况二
return [-1, -1]
```
```python
# 解法1的代码简化,直接使用left和right值作为左右边界标记
class Solution:
def searchRange(self, nums: List[int], target: int) -> List[int]:
# 查找左边界(第一个等于 target 的位置)
def getLeftBorder(nums: List[int], target: int) -> int:
left, right = 0, len(nums) - 1
while left <= right:
middle = left + (right - left) // 2
if nums[middle] >= target: # nums[middle] >= target 时,继续向左查找
right = middle - 1
else:
left = middle + 1
return left

# 查找右边界(最后一个等于 target 的位置)
def getRightBorder(nums: List[int], target: int) -> int:
left, right = 0, len(nums) - 1
while left <= right:
middle = left + (right - left) // 2
if nums[middle] > target: # nums[middle] > target 时,继续向右查找
right = middle - 1
else:
left = middle + 1
return right

# 查找左边界和右边界
leftBorder = getLeftBorder(nums, target)
rightBorder = getRightBorder(nums, target)

# 判断目标值是否存在
if leftBorder <= rightBorder:
return [leftBorder, rightBorder]
else:
return [-1, -1]
```
```python
# 解法2
# 1、首先,在 nums 数组中二分查找 target;
# 2、如果二分查找失败,则 binarySearch 返回 -1,表明 nums 中没有 target。此时,searchRange 直接返回 {-1, -1};
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25 changes: 24 additions & 1 deletion problems/0203.移除链表元素.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -367,9 +367,32 @@ class Solution {
```

### Python:
用原来的链表操作:
```python
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
while head != None and head.val == val:
head = head.next
cur = head
while cur != None and cur.next != None:
check = cur.next
if check.val == val:
cur.next = check.next
check = check.next
else:
cur = cur.next
check = check.next
return head

```

```python
(版本一)虚拟头节点法
虚拟头节点法
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
Expand Down

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