12/3
1、Two Sum
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
class Solution:
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
dict = {}
for i in range(len(nums)):
j = target - nums[i]
if j in dict:
return [dict[j],i]
else:
dict[nums[i]] = i
#return 0
2. Add Two Numbers
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
class Solution:
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
ans = 0
unit = 1
while l1 or l2:
if l1:
ans += l1.val * unit
l1 = l1.next
if l2:
ans += l2.val * unit
l2 = l2.next
unit *= 10 alpha = cur = ListNode(0) for n in reversed(str(ans)):
cur.next = ListNode(int(n))
cur = cur.next return alpha.next
补充：链表是由一些节点构成的，这些节点之间由指针连接，形成了一个链式结构。最基本的链表节点只需要存储当前节点的值，和一个指向下一节点的指针。由这种只存储下一节点地址的链表节点构成的链表被称为单向链表。
在节点ListNode定义中，定义为节点为结构变量。节点存储了两个变量：value 和 next。value 是这个节点的值，next 是指向下一节点的指针，当 next 为空指针时，这个节点是链表的最后一个节点。构造函数包含两个参数 _value 和 _next ，分别用来给节点赋值和指定下一节点。
struct ListNode {
int val; //定义val变量值，存储节点值
struct ListNode *next; //定义next指针，指向下一个节点，维持节点连接
}
203. Remove Linked List Elements( 类比 83. Remove Duplicates from Sorted List)
Input: 1->2->6->3->4->5->6, val = 6
Output: 1->2->3->4->5
class Solution:
def removeElements(self, head, val):
"""
:type head: ListNode
:type val: int
:rtype: ListNode
"""
if head == None:
return head
dummy = ListNode(0)
dummy.next = head
pre = dummy
while head:
if head.val == val:
pre.next = head.next
head = pre
pre = head
head = head.next
return dummy.next
（链表）206. Reverse Linked List
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL
class Solution:
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
tail = None
cur = head
while cur:
next_node = cur.next #存储当前节点的指向
cur.next = tail #当前节点指向尾节点，（改变指针指向）
tail = cur
cur = next_node
return tail
21. Merge Two Sorted Lists
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
class Solution:
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
head = ListNode(0)
cur = head while l1 and l2: if l1.val > l2.val:
cur.next = l2
l2 = l2.next else:
cur.next = l1
l1 = l1.next cur = cur.next cur.next = l1 or l2 return head.next
26. Remove Duplicates from Sorted Array
Given nums = [0,0,1,1,1,2,2,3,3,4], Your function should return length =5, with the first five elements ofnumsbeing modified to0,1,2,3, and4respectively. It doesn't matter what values are set beyond the returned length.
class Solution:
def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
cnt=0
if (len(nums)==0):
return cnt for i in sorted(set(nums)):
#set() 函数创建一个无序不重复元素集，可进行关系测试，删除重复数据，还可以计算交集、差集、并集等。
nums[cnt]=i
cnt+=1 return cnt
27. Remove Element
Given nums = [0,1,2,2,3,0,4,2], val = 2, Your function should return length =5, with the first five elements ofnumscontaining0,1,3,0, and 4.
class Solution:
def removeElement(self, nums, val):
"""
:type nums: List[int]
:type val: int
:rtype: int
"""
index = 0
while(index < len(nums)):
if (nums[index] == val):
nums.pop(nums.index(val))
#pop()指定删除对象的索引位置，例如，a.pop(3)要删除列表a中索引3对应的元素。
else:
index += 1
88. Merge Sorted Array
Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6], n = 3 Output: [1,2,2,3,5,6]
class Solution:
def merge(self, nums1, m, nums2, n):
"""
:type nums1: List[int]
:type m: int
:type nums2: List[int]
:type n: int
:rtype: void Do not return anything, modify nums1 in-place instead.
"""
for v in nums2:
nums1[m] = v
m+=1
nums1.sort()
(二叉树）111. Minimum Depth of Binary Tree
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its minimum depth = 2.
class Solution:
def minDepth(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root:
return 0 children = [root.left, root.right]
# if we're at leaf node
if not any(children):
return 1 min_depth = float('inf')
for c in children:
if c:
min_depth = min(self.minDepth(c), min_depth)
return min_depth + 1
12/4
563. Binary Tree Tilt
Example:
Input:
1
/ \
2 3
Output: 1
Explanation:
Tilt of node 2 : 0
Tilt of node 3 : 0
Tilt of node 1 : |2-3| = 1
Tilt of binary tree : 0 + 0 + 1 = 1
class Solution:
def findTilt(self, root):
"""
:type root: TreeNode
:rtype: int
"""
self.res = []
if not root:
return 0
self.dfs(root)
return sum(self.res)
def dfs(self,root):
if not root:
return 0
if not root.left and not root.right:
return root.val
leftsum = self.dfs(root.left)
rightsum = self.dfs(root.right)
self.res.append(abs(leftsum - rightsum))
return root.val + leftsum + rightsum
102. 二叉树的层次遍历
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
class Solution(object):
def levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
# write code here
if not root:
return []
queue=[root]
outList=[]
while queue:
res=[]
nextQueue=[]
for point in queue: #这里再遍历每一层
res.append(point.val)
if point.left:
nextQueue.append(point.left)
if point.right:
nextQueue.append(point.right)
outList.append(res)
queue=nextQueue #这一步很巧妙，用当前层覆盖上一层
return outList
83. Remove Duplicates from Sorted List
Input: 1->1->2->3->3
Output: 1->2->3
class Solution:
def deleteDuplicates(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if head is None:
return head
cur = head
while cur.next:
if cur.val == cur.next.val:
cur.next = cur.next.next
continue
cur = cur.next
return head
237. Delete Node in a Linked List
4 -> 5 -> 1 -> 9
Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None class Solution:
def deleteNode(self, node):
"""
:type node: ListNode
:rtype: void Do not return anything, modify node in-place instead.
"""
node.val = node.next.val
node.next = node.next.next
35. Search Insert Position
Input: [1,3,5,6], 5
Output: 2
Input: [1,3,5,6], 2
Output: 1
class Solution:
def searchInsert(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
index = 0
for i in nums:
diff = target - i
if diff > 0:
index += 1
else:
break
return index
724. Find Pivot Index
Input:
nums = [1, 7, 3, 6, 5, 6]
Output: 3
Explanation:
The sum of the numbers to the left of index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3.
Also, 3 is the first index where this occurs.
class Solution:
def pivotIndex(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
lefty, righty = 0, sum(nums[1:])
if nums and lefty == righty:
return 0
for i in range(1, len(nums)):
lefty += nums[i-1]
righty -= nums[i]
if lefty == righty:
return i
return -1
674. Longest Continuous Increasing Subsequence
Input: [1,3,5,4,7]
Output: 3
class Solution:
def findLengthOfLCIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if not nums:
return 0
dp = [1] * len(nums)
for i in range(1, len(nums)):
if nums[i] > nums[i-1]:
dp[i] = dp[i - 1] + 1
return max(dp)
#返回最长序列的起始索引
#return dp.index(max(dp))-max(dp)+1
108. Convert Sorted Array to Binary Search Tree
class Solution:
def sortedArrayToBST(self, nums):
"""
:type nums: List[int]
:rtype: TreeNode
"""
if not nums:
return None mid = len(nums) // 2
#" // "来表示整数除法，返回不大于结果的一个最大的整数，而" / " 则单纯的表示浮点数除法 root = TreeNode(nums[mid])
root.left = self.sortedArrayToBST(nums[:mid])
root.right = self.sortedArrayToBST(nums[mid+1:]) return root
169. Majority Element
Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
Input: [3,2,3]
Output: 3
class Solution:
def majorityElement(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
a = int(len(nums)/2)
b = collections.Counter(nums)
for i in nums:
if b[i]>a:
return i
242. Valid Anagram
Input: s = "anagram", t = "nagaram"
Output: true
class Solution:
def isAnagram(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
return collections.Counter(s) == collections.Counter(t)
#Counter（计数器）是对字典的补充，用于追踪值的出现次数。
283. Move Zeroes
Input:[0,1,0,3,12]
Output:[1,3,12,0,0]
class Solution:
def moveZeroes(self, nums):
"""
:type nums: List[int]
:rtype: void Do not return anything, modify nums in-place instead.
"""
length = len(nums)
i = 0
while i < length: if nums[i] == 0:
nums.append(nums.pop(i))
length -= 1
continue i += 1
268. Missing Number
Input: [9,6,4,2,3,5,7,0,1]
Output: 8
class Solution:
def missingNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
length=len(nums)
return (int((length**2+length)/2))-sum(nums)
387. First Unique Character in a String
s = "loveleetcode",
return 2.
class Solution:
def firstUniqChar(self, s):
"""
:type s: str
:rtype: int
"""
hash_table = {e:index for index, e in enumerate(s)} for index,e in enumerate(s):
if e in hash_table:
if hash_table[e]==index:
return index
del hash_table[e]
return -1
13. Roman to Integer
Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
Input: "III"
Output: 3
Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.
class Solution:
def romanToInt(self, s):
"""
:type s: str
:rtype: int
"""
dic = {'M':1000,
'D':500,
'C':100,
'L':50,
'X':10,
'V':5,
'I':1} temp = 0
res = 0
for c in s:
if dic[c] > temp:
res -= 2 * temp
temp = dic[c]
res += temp return res
350. Intersection of Two Arrays II
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]
class Solution:
def intersect(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
counts = collections.Counter(nums1)
res = [] for num in nums2:
if counts[num] > 0:
res.append(num)
counts[num] -= 1 return res
2019/3/29
28. 实现strStr()
（如何判断一个字符串是否包含另一个字符串）
class Solution(object):
def strStr(self, haystack, needle):
"""
:type haystack: str
:type needle: str
:rtype: int
"""
m = len(haystack)
n = len(needle)
for i in range(m-n+1):
if haystack[i:i+n] == needle:
return i
return -1
26. 删除排序数组中的重复项
class Solution(object):
def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
n = len(nums) j = 0
for i in range(n-1):
if nums[i] == nums[i+1]:
nums[j] = nums[i] else:
nums[j] = nums[i]
nums[j+1] = nums[i+1]
j = j+1 return j+1
35. 搜索插入位置
class Solution(object):
def searchInsert(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
n = len(nums) if target > nums[n-1]:
return n
elif target < nums[0]:
return 0
for i in range(n):
while target == nums[i]:
return i
while target > nums[i] and target < nums[i+1]:
return i+1
class Solution(object):
def reverse(self, x):
"""
:type x: int
:rtype: int
"""
string = str(x)
if x < 0:
string = "-" + string[len(string):0:-1].lstrip("")
elif x > 0:
string = string[::-1].lstrip("")
else:
string = ""
if -2**31<int(string)<2**31-1:
return int(string)
else:
return 0