leetcode简单(161-180)python
690. Employee Importance(e-161)
You are given a data structure of employee information, which includes the employee's unique id, his importance value and his direct subordinates' id.
For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. They have importance value 15, 10 and 5, respectively. Then employee 1 has a data structure like [1, 15, [2]], and employee 2 has [2, 10, [3]], and employee 3 has [3, 5, []]. Note that although employee 3 is also a subordinate of employee 1, the relationship is not direct.
Now given the employee information of a company, and an employee id, you need to return the total importance value of this employee and all his subordinates.
class Employee(object):
def __init__(self, id, importance, subordinates):
# It's the unique id of each node.
# unique id of this employee
self.id = id
# the importance value of this employee
self.importance = importance
# the id of direct subordinates
self.subordinates = subordinates
class Solution(object):
def getImportance(self, employees, id):
"""
:type employees: Employee
:type id: int
:rtype: int
"""
# Time: O(n)
# Space: O(n)
emps = {employee.id: employee for employee in employees}
def dfs(id):
subordinates_importance = sum([dfs(sub_id) for sub_id in emps[id].subordinates])
return subordinates_importance + emps[id].importance
return dfs(id)
693. Binary Number with Alternating Bits(e-162)
Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.
class Solution(object):
def hasAlternatingBits(self, n):
"""
:type n: int
:rtype: bool
"""
flag=n%2
n=n//2
while n>0:
if n%2==flag:
return False
flag=n%2
n=n//2
return True695. Max Area of Island(e-163)
Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.
Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)
class Solution(object):
def maxAreaOfIsland(self, grid):
m, n = len(grid), len(grid[0])
def dfs(i, j):
if 0 <= i < m and 0 <= j < n and grid[i][j]:
grid[i][j] = 0
return 1 + dfs(i - 1, j) + dfs(i, j + 1) + dfs(i + 1, j) + dfs(i, j - 1)
return 0
areas = [dfs(i, j) for i in range(m) for j in range(n) if grid[i][j]]
return max(areas) if areas else 0
696. Count Binary Substrings(e-164)
Give a string s, count the number of non-empty (contiguous) substrings that have the same number of 0's and 1's, and all the 0's and all the 1's in these substrings are grouped consecutively.
Substrings that occur multiple times are counted the number of times they occur.
class Solution(object):
def countBinarySubstrings(self, s):
"""
:type s: str
:rtype: int
"""
group=[1]
for i in range(1,len(s)):
if s[i-1]!=s[i]:
group.append(1)
else:
group[-1]+=1
ans=0
for i in range(1, len(group)):
ans+=min(group[i-1], group[i])
return ans
697. Degree of an Array(e-165)
Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.
Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums
class Solution(object):
def findShortestSubArray(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
#定义字典元素以及对应的位置信息，count记录每个元素出现的次数
left, right, count = {}, {}, {}
for i, num in enumerate(nums):
#如果元素num第一次出现，则存入left中
if num not in left:
left[num] = i
#将出现的元素及对应的位置信息存入right中
right[num] = i
#每个元素出现一次，则count对位num位置的数值 +1
count[num] = count.get(num, 0) + 1
ans = len(nums)
degree = max(count.values())
for num in count:
if count[num]==degree:
ans = min(ans, right[num]-left[num]+1)
return ans
700. Search in a Binary Search Tree(e-166)---------------
Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL.
703. Kth Largest Element in a Stream(e-167)--------------
Design a class to find the kth largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth distinct element.
Your KthLargest class will have a constructor which accepts an integer k and an integer array nums, which contains initial elements from the stream. For each call to the method KthLargest.add, return the element representing the kth largest element in the stream.
704. Binary Search(e-168)
Given a sorted (in ascending order) integer array nums of n elements and a target value, write a function to search targetin nums. If target exists, then return its index, otherwise return -1.
class Solution:
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == target:
return mid
elif nums[mid] < target:
left = mid + 1
else:
right = mid - 1
return -1
705. Design HashSet(e-169)-------
Design a HashSet without using any built-in hash table libraries.
To be specific, your design should include these two functions:
add(value): Insert a value into the HashSet.contains(value): Return whether the value exists in the HashSet or not.remove(value): Remove a value in the HashSet. If the value does not exist in the HashSet, do nothing.
706. Design HashMap(e-170)-------------
Design a HashMap without using any built-in hash table libraries.
To be specific, your design should include these two functions:
put(key, value): Insert a (key, value) pair into the HashMap. If the value already exists in the HashMap, update the value.get(key): Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.remove(key): Remove the mapping for the value key if this map contains the mapping for the key.
707. Design Linked List(e-171)---------------
Design your implementation of the linked list. You can choose to use the singly linked list or the doubly linked list. A node in a singly linked list should have two attributes: val and next. val is the value of the current node, and next is a pointer/reference to the next node. If you want to use the doubly linked list, you will need one more attribute prev to indicate the previous node in the linked list. Assume all nodes in the linked list are 0-indexed.
Implement these functions in your linked list class:
- get(index) : Get the value of the
index-th node in the linked list. If the index is invalid, return-1. - addAtHead(val) : Add a node of value
valbefore the first element of the linked list. After the insertion, the new node will be the first node of the linked list. - addAtTail(val) : Append a node of value
valto the last element of the linked list. - addAtIndex(index, val) : Add a node of value
valbefore theindex-th node in the linked list. Ifindexequals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted. - deleteAtIndex(index) : Delete the
index-th node in the linked list, if the index is valid.
709. To Lower Case(e-172)
Implement function ToLowerCase() that has a string parameter str, and returns the same string in lowercase.
class Solution(object):
def toLowerCase(self, str):
"""
:type str: str
:rtype: str
"""
return str.lower()717. 1-bit and 2-bit Characters(e-173)
class Solution(object):
def isOneBitCharacter(self, bits):
"""
:type bits: List[int]
:rtype: bool
"""
i=0
while i 720. Longest Word in Dictionary(e-174)
Given a list of strings words representing an English Dictionary, find the longest word in words that can be built one character at a time by other words in words. If there is more than one possible answer, return the longest word with the smallest lexicographical order.
class Solution(object):
def longestWord(self, words):
"""
:type words: List[str]
:rtype: str
"""
wset = set([''])
ans = ''
for word in sorted(words):
if word[:-1] in wset:
wset.add(word)
if len(word) > len(ans):
ans = word
return ans
724. Find Pivot Index(e-175)
Given an array of integers nums, write a method that returns the "pivot" index of this array.
We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the index.
If no such index exists, we should return -1. If there are multiple pivot indexes, you should return the left-most pivot index.
class Solution(object):
def pivotIndex(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
ret=sum(nums)
left=0
for i,k in enumerate(nums):
if left*2+k==vet:
return k
left+=k
return -1
728. Self Dividing Numbers(e-176)
A self-dividing number is a number that is divisible by every digit it contains.
For example, 128 is a self-dividing number because 128 % 1 == 0, 128 % 2 == 0, and 128 % 8 == 0.
Also, a self-dividing number is not allowed to contain the digit zero.
Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible.
class Solution(object):
def selfDividingNumbers(self, left, right):
"""
:type left: int
:type right: int
:rtype: List[int]
"""
is_self_dividing=lambda num:'0' not in str(num) and all([num%int(di)==0 for di in str(num)])
return filter(is_self_dividing, range(left, right+1))733. Flood Fill(e-177)
An image is represented by a 2-D array of integers, each integer representing the pixel value of the image (from 0 to 65535).
Given a coordinate (sr, sc) representing the starting pixel (row and column) of the flood fill, and a pixel value newColor, "flood fill" the image.
To perform a "flood fill", consider the starting pixel, plus any pixels connected 4-directionally to the starting pixel of the same color as the starting pixel, plus any pixels connected 4-directionally to those pixels (also with the same color as the starting pixel), and so on. Replace the color of all of the aforementioned pixels with the newColor.
At the end, return the modified image.
class Solution(object):
def floodFill(self, image, sr, sc, newColor):
"""
:type image: List[List[int]]
:type sr: int
:type sc: int
:type newColor: int
:rtype: List[List[int]]
"""
SR, SC = len(image), len(image[0])
color = image[sr][sc]
if color == newColor: return image
def dfs(r, c):
if image[r][c] == color:
image[r][c] = newColor
if r >= 1: dfs(r - 1, c)
if r < SR - 1: dfs(r + 1, c)
if c >= 1: dfs(r, c - 1)
if c < SC - 1: dfs(r, c + 1)
dfs(sr, sc)
return image
744. Find Smallest Letter Greater Than Target(e-178)
Given a list of sorted characters letters containing only lowercase letters, and given a target letter target, find the smallest element in the list that is larger than the given target.
Letters also wrap around. For example, if the target is target = 'z' and letters = ['a', 'b'], the answer is 'a'.
class Solution(object):
def nextGreatestLetter(self, letters, target):
"""
:type letters: List[str]
:type target: str
:rtype: str
"""
for letter in letters:
if letter>target:
return letter
return letters[0]746. Min Cost Climbing Stairs(e-179)
On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).
Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step with index 0, or the step with index 1.
class Solution(object):
def minCostClimbingStairs(self, cost):
"""
:type cost: List[int]
:rtype: int
"""
dp0, dp1,dp2=0,0,0
for i in range(2, len(cost)+1):
dp2=min(dp0+cost[i-2], dp1+cost[i-1])
dp0, dp1=dp1,dp2
return dp2747. Largest Number At Least Twice of Others(e-180)
In a given integer array nums, there is always exactly one largest element.
Find whether the largest element in the array is at least twice as much as every other number in the array.
If it is, return the index of the largest element, otherwise return -1.
class Solution(object):
def dominantIndex(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums)==1:
return 0
num=sorted(nums)
if num[-1]>=num[-2]*2:
return nums.index(num[-1])
return -1