链表相加
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
s1, s2 = [], []
while l1:
s1.append(l1.val)
l1 = l1.next
while l2:
s2.append(l2.val)
l2 = l2.next
ans = None
carry = 0
while s1 or s2 or carry != 0:
a = 0 if not s1 else s1.pop()
b = 0 if not s2 else s2.pop()
cur = a + b + carry
carry = cur // 10
cur %= 10
curnode = ListNode(cur)
curnode.next = ans
ans = curnode
return ans
两数相加
def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
pre = ListNode(0)
cur = pre
carry = 0
while l1 or l2:
x = l1.val if l1 else 0
y = l2.val if l2 else 0
sum_v = x + y + carry
carry = sum_v // 10
sum_v = sum_v % 10
cur.next = ListNode(sum_v)
cur = cur.next
if l1:
l1 = l1.next
if l2:
l2 = l2.next
if carry == 1:
cur.next = ListNode(carry)
return pre.next
有序数组中位数
def findMedianSortedArrays(nums1: List[int], nums2: List[int]) -> float:
def getKthElement(k):
index1, index2 = 0, 0
while True:
# 特殊情况
if index1 == m:
return nums2[index2 + k - 1]
if index2 == n:
return nums1[index1 + k - 1]
if k == 1:
return min(nums1[index1], nums2[index2])
# 正常情况
newIndex1 = min(index1 + k // 2 - 1, m - 1)
newIndex2 = min(index2 + k // 2 - 1, n - 1)
pivot1, pivot2 = nums1[newIndex1], nums2[newIndex2]
if pivot1 <= pivot2:
k -= newIndex1 - index1 + 1
index1 = newIndex1 + 1
else:
k -= newIndex2 - index2 + 1
index2 = newIndex2 + 1
m, n = len(nums1), len(nums2)
totalLength = m + n
if totalLength % 2 == 1:
return getKthElement((totalLength + 1) // 2)
else:
return (getKthElement(totalLength // 2) + getKthElement(totalLength // 2 + 1)) / 2
数组中逆序对数
class Solution:
def mergeSort(self, nums, tmp, l, r):
if l >= r:
return 0
mid = (l + r) // 2
inv_count = self.mergeSort(nums, tmp, l, mid) + self.mergeSort(nums, tmp, mid + 1, r)
i, j, pos = l, mid + 1, l
while i <= mid and j <= r:
if nums[i] <= nums[j]:
tmp[pos] = nums[i]
i += 1
inv_count += (j - (mid + 1))
else:
tmp[pos] = nums[j]
j += 1
pos += 1
for k in range(i, mid + 1):
tmp[pos] = nums[k]
inv_count += (j - (mid + 1))
pos += 1
for k in range(j, r + 1):
tmp[pos] = nums[k]
pos += 1
nums[l:r+1] = tmp[l:r+1]
return inv_count
def reversePairs(self, nums: List[int]) -> int:
n = len(nums)
tmp = [0] * n
return self.mergeSort(nums, tmp, 0, n - 1)
之字打印二叉树
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if not root: return []
res,deque = [],collections.deque([root])
while deque:
tmp = collections.deque()
for _ in range(len(deque)):
node = deque.popleft()
if len(res) % 2 : tmp.appendleft(node.val)
else:tmp.append(node.val)
if node.left:deque.append(node.left)
if node.right: deque.append(node.right)
res.append(list(tmp))
return res
数值的整数次方
class Solution:
def myPow(self, x: float, n: int) -> float:
if x == 0 : return 0
if n < 0 : x,n = 1/x,-n
res = 1
while n:
if n & 1 : res *= x
x *= x
n >>= 1
return res
单词拆分
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
l = len(s)
if not wordDict: return not s
dp = [False] * (l + 1)
dp[0] = True
for i in range(1, l + 1):
for j in range(i - 1, -1, -1):
if dp[j] and s[j:i] in wordDict:
dp[i] = True
break
return dp[-1]
接雨水
class Solution:
def trap(self, height: List[int]) -> int:
if not height:
return 0
n = len(height)
leftMax = [height[0]] + [0] * (n - 1)
for i in range(1, n):
leftMax[i] = max(leftMax[i - 1], height[i])
rightMax = [0] * (n - 1) + [height[n - 1]]
for i in range(n - 2, -1, -1):
rightMax[i] = max(rightMax[i + 1], height[i])
ans = sum(min(leftMax[i], rightMax[i]) - height[i] for i in range(n))
return ans
最长重复子数组
class Solution:
def findLength(self, A: List[int], B: List[int]) -> int:
n,m = len(A),len(B)
dp = [[0] * (m+1) for _ in range(n+1)]
ans = 0
for i in range(n-1,-1,-1):
for j in range(m-1,-1,-1):
dp[i][j] = dp[i+1][j+1] + 1 if A[i] == B[j] else 0
ans = max(ans,dp[i][j])
return ans
0-1缺失的数字
class Solution:
def missingNumber(self, nums: List[int]) -> int:
left = 0
right = len(nums)-1
while left <= right:
mid = left + (right - left ) // 2
if nums[mid] == mid :
left = mid+1
else:
right = mid -1
return left
路径总和
class Solution:
def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]:
res = []
path = []
def dfs(root,targetSum):
if not root:
return
path.append(root.val)
targetSum -= root.val
if not root.left and not root.right and targetSum == 0:
res.append(path[:])
dfs(root.left,targetSum)
dfs(root.right,targetSum)
path.pop()
dfs(root,targetSum)
return res
判断两棵树是否相同
class Solution:
def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
if not p and not q:
return True
if not p or not q:
return False
if p.val != q.val:
return False
return self.isSameTree(p.right,q.right) and self.isSameTree(p.left,q.left)
二叉搜索树最近祖先
class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if p.val > root.val and q.val > root.val:
return self.lowestCommonAncestor(root.right,p,q)
elif p.val < root.val and q.val < root.val:
return self.lowestCommonAncestor(root.left,p,q)
else:
return root
二叉树镜像
class Solution:
def mirrorTree(self, root: TreeNode) -> TreeNode:
if not root: return
stack = [root]
while stack:
node = stack.pop()
if node.left:stack.append(node.left)
if node.right:stack.append(node.right)
node.left,node.right = node.right,node.left
return root
class Solution:
def mirrorTree(self, root: TreeNode) -> TreeNode:
if not root: return
root.left, root.right = self.mirrorTree(root.right), self.mirrorTree(root.left)
return root
二叉树层次遍历
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
res = []
if not root:
return res
queue = [root]
while queue:
n = len(queue)
tmp = []
for _ in range(n):
r = queue.pop(0)
tmp.append(r.val)
if r.left:
queue.append(r.left)
if r.right:
queue.append(r.right)
res.append(tmp)
return res
二叉树前序遍历
class Solution:
def preorderTraversal(self, root: TreeNode) -> List[int]:
stack,rst = [root],[]
while stack:
i = stack.pop()
if isinstance(i,TreeNode):
stack.extend([i.right,i.left,i.val])
elif isinstance(i,int):
rst.append(i)
return rst
class Solution:
def preorderTraversal(self, root: TreeNode) -> List[int]:
def preorder(root: TreeNode):
if not root:
return
res.append(root.val)
preorder(root.left)
preorder(root.right)
res = list()
preorder(root)
return res
