diff --git a/problems/0129.求根到叶子节点数字之和.md b/problems/0129.求根到叶子节点数字之和.md index b37270e2..17642793 100644 --- a/problems/0129.求根到叶子节点数字之和.md +++ b/problems/0129.求根到叶子节点数字之和.md @@ -165,7 +165,32 @@ public: Java: Python: +```python3 +class Solution: + def sumNumbers(self, root: TreeNode) -> int: + res = 0 + path = [] + def backtrace(root): + nonlocal res + if not root: return # 节点空则返回 + path.append(root.val) + if not root.left and not root.right: # 遇到了叶子节点 + res += get_sum(path) + if root.left: # 左子树不空 + backtrace(root.left) + if root.right: # 右子树不空 + backtrace(root.right) + path.pop() + def get_sum(arr): + s = 0 + for i in range(len(arr)): + s = s * 10 + arr[i] + return s + + backtrace(root) + return res +``` Go: JavaScript: diff --git a/problems/0143.重排链表.md b/problems/0143.重排链表.md index 62232051..76df63b7 100644 --- a/problems/0143.重排链表.md +++ b/problems/0143.重排链表.md @@ -222,7 +222,61 @@ public class ReorderList { ``` Python: +```python3 +# 方法二 双向队列 +class Solution: + def reorderList(self, head: ListNode) -> None: + """ + Do not return anything, modify head in-place instead. + """ + d = collections.deque() + tmp = head + while tmp.next: # 链表除了首元素全部加入双向队列 + d.append(tmp.next) + tmp = tmp.next + tmp = head + while len(d): # 一后一前加入链表 + tmp.next = d.pop() + tmp = tmp.next + if len(d): + tmp.next = d.popleft() + tmp = tmp.next + tmp.next = None # 尾部置空 + +# 方法三 反转链表 +class Solution: + def reorderList(self, head: ListNode) -> None: + if head == None or head.next == None: + return True + slow, fast = head, head + while fast and fast.next: + slow = slow.next + fast = fast.next.next + right = slow.next # 分割右半边 + slow.next = None # 切断 + right = self.reverseList(right) #反转右半边 + left = head + # 左半边一定比右半边长, 因此判断右半边即可 + while right: + curLeft = left.next + left.next = right + left = curLeft + curRight = right.next + right.next = left + right = curRight + + + def reverseList(self, head: ListNode) -> ListNode: + cur = head + pre = None + while(cur!=None): + temp = cur.next # 保存一下cur的下一个节点 + cur.next = pre # 反转 + pre = cur + cur = temp + return pre +``` Go: JavaScript: diff --git a/problems/0234.回文链表.md b/problems/0234.回文链表.md index 6a24b1d0..b3ad899c 100644 --- a/problems/0234.回文链表.md +++ b/problems/0234.回文链表.md @@ -148,7 +148,62 @@ public: ## Python -```python +```python3 +#数组模拟 +class Solution: + def isPalindrome(self, head: ListNode) -> bool: + length = 0 + tmp = head + while tmp: #求链表长度 + length += 1 + tmp = tmp.next + + result = [0] * length + tmp = head + index = 0 + while tmp: #链表元素加入数组 + result[index] = tmp.val + index += 1 + tmp = tmp.next + + i, j = 0, length - 1 + while i < j: # 判断回文 + if result[i] != result[j]: + return False + i += 1 + j -= 1 + return True + +#反转后半部分链表 +class Solution: + def isPalindrome(self, head: ListNode) -> bool: + if head == None or head.next == None: + return True + slow, fast = head, head + while fast and fast.next: + pre = slow + slow = slow.next + fast = fast.next.next + + pre.next = None # 分割链表 + cur1 = head # 前半部分 + cur2 = self.reverseList(slow) # 反转后半部分,总链表长度如果是奇数,cur2比cur1多一个节点 + while cur1: + if cur1.val != cur2.val: + return False + cur1 = cur1.next + cur2 = cur2.next + return True + + def reverseList(self, head: ListNode) -> ListNode: + cur = head + pre = None + while(cur!=None): + temp = cur.next # 保存一下cur的下一个节点 + cur.next = pre # 反转 + pre = cur + cur = temp + return pre ``` ## Go diff --git a/problems/0724.寻找数组的中心索引.md b/problems/0724.寻找数组的中心索引.md index 3ed68d47..b4115893 100644 --- a/problems/0724.寻找数组的中心索引.md +++ b/problems/0724.寻找数组的中心索引.md @@ -89,7 +89,16 @@ class Solution { ## Python -```python +```python3 +class Solution: + def pivotIndex(self, nums: List[int]) -> int: + numSum = sum(nums) #数组总和 + leftSum = 0 + for i in range(len(nums)): + if numSum - leftSum -nums[i] == leftSum: #左右和相等 + return i + leftSum += nums[i] + return -1 ``` ## Go diff --git a/problems/0922.按奇偶排序数组II.md b/problems/0922.按奇偶排序数组II.md index 92db204d..97d7091e 100644 --- a/problems/0922.按奇偶排序数组II.md +++ b/problems/0922.按奇偶排序数组II.md @@ -149,7 +149,32 @@ class Solution { ## Python -```python +```python3 +#方法2 +class Solution: + def sortArrayByParityII(self, nums: List[int]) -> List[int]: + result = [0]*len(nums) + evenIndex = 0 + oddIndex = 1 + for i in range(len(nums)): + if nums[i] % 2: #奇数 + result[oddIndex] = nums[i] + oddIndex += 2 + else: #偶数 + result[evenIndex] = nums[i] + evenIndex += 2 + return result + +#方法3 +class Solution: + def sortArrayByParityII(self, nums: List[int]) -> List[int]: + oddIndex = 1 + for i in range(0,len(nums),2): #步长为2 + if nums[i] % 2: #偶数位遇到奇数 + while nums[oddIndex] % 2: #奇数位找偶数 + oddIndex += 2 + nums[i], nums[oddIndex] = nums[oddIndex], nums[i] + return nums ``` ## Go