39
Title description:
Given an array of candidates with no repeated elements and a target number target, find out all combinations of candidates that can make the sum of numbers target.
The numbers in candidates can be repeatedly selected without limitation.
Example:
Answer:
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
candidates.sort()
n = len(candidates)
res = []
def helper(i, tmp_sum, tmp):
if tmp_sum > target or i == n:
return
if tmp_sum == target:
res.append(tmp)
return
helper(i, tmp_sum + candidates[i],tmp + [candidates[i]])
helper(i+1, tmp_sum ,tmp)
helper(0, 0, [])
return res
224
Title description:
Implement a basic calculator to calculate the value of a simple string expression s.
Example:
Answer:
class Solution:
def calculate(self, s: str) -> int:
#将“(”前的和加进stack中，将“)”前的数从stack中取出和前面的和相加
stack = []
operator = 1
res = 0
num = 0
for i in range(len(s)):
if s[i] == '(':
stack.append(res)
stack.append(operator)
res = 0
operator = 1
num = 0
elif s[i] == ')':
res = res + num*operator
operator = stack.pop()
res = stack.pop() + res*operator
num = 0
operator = 1
elif s[i] == '+':
res = res + num * operator
num = 0
operator = 1
elif s[i] == '-':
res = res + num * operator
num = 0
operator = -1
elif s[i] != ' ':
num = num*10 + int(s[i])
res = res + num * operator
return res
227
Title description:
Implement a basic calculator to calculate the value of a simple string expression.
String expressions only include non-negative integers, +, -, *, / four operators and spaces. Integer division only retains the integer part.
Example:
Answer:
class Solution:
def calculate(self, s: str) -> int:
sign='+'
s+="+"
stack=[]
num=0
for ch in s:
if ch>='0' and ch <='9':
num=10*num+int(ch)
continue
if ch==" ":
continue
if sign=='+':
stack.append(num)
if sign=='-':
stack.append(-num)
if sign=="*":
pre=stack.pop()
stack.append(pre*num)
if sign=='/':
pre=stack.pop()
stack.append(int(pre/num))
sign=ch
num=0
return sum(stack)
576
Title description:
Given two strings s1 and s2, write a function to determine whether s2 contains the permutation of s1.
In other words, one of the permutations of the first string is a substring of the second string.
Example:
Answer:
class Solution:
def checkInclusion(self, s1: str, s2: str) -> bool:
l1, l2 = len(s1), len(s2)
c1 = collections.Counter(s1)
c2 = collections.Counter()
cnt = 0 #统计变量，全部26个字符，频率相同的个数，当cnt==s1字母的个数的时候，就是全部符合题意，返回真
p = q = 0 #滑动窗口[p,q]
while q < l2:
c2[s2[q]] += 1
if c1[s2[q]] == c2[s2[q]]: #对于遍历到的字母，如果出现次数相同
cnt += 1 #统计变量+1
if cnt == len(c1): #判断结果写在前面，此时证明s2滑动窗口和s1全部字符相同，返回真
return True
q += 1 #滑动窗口右移
if q - p + 1 > l1: #这是构造第一个滑动窗口的特殊判断，里面内容是维护边界滑动窗口
if c1[s2[p]] == c2[s2[p]]: #判断性的if写在前面，因为一旦频率变化，这个统计变量就减1
cnt -= 1
c2[s2[p]] -= 1 #字典哈希表移除最前面的字符
if c2[s2[p]] == 0: #由于counter特性，如果value为0，必须删除它
del c2[s2[p]]
p += 1 #最前面的下标右移动
return False
703
Title description:
Design a class that finds the k-th largest element in the data stream. Note that it is the k-th largest element after sorting, not the k-th different element.
Please implement the KthLargest class:
KthLargest(int k, int[] nums) Initialize the object using integer k and integer stream nums.
int add(int val) After inserting val into the data stream nums, return the k-th largest element in the current data stream.
Example:
Answer:
class KthLargest:
#最小堆
def __init__(self, k: int, nums: List[int]):
self.heap = []
self.k = k
for num in nums:
heapq.heappush(self.heap,num)
if len(self.heap) > k:
heapq.heappop(self.heap)
def add(self, val: int) -> int:
heapq.heappush(self.heap,val)
if len(self.heap) > self.k:
heapq.heappop(self.heap)
return self.heap[0]
# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)