【算法】力扣第 265场周赛
作者:互联网
5914. 值相等的最小索引
送分题,数据量也小,才100
class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for i,n in enumerate(nums):
if i%10==n:return i
return -1
一行:
class Solution:
def smallestEqual(self, nums: List[int]) -> int:
return ([i for i,n in enumerate(nums) if i%10==n]+[-1])[0]
5915. 找出临界点之间的最小和最大距离
连续记录3个结点,为临界点时记录位置(cnt),最后再找一下距离
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def nodesBetweenCriticalPoints(self, head: Optional[ListNode]) -> List[int]:
if not head.next.next:return [-1,-1]
a,b,c=head,head.next,head.next.next
res=[]
cnt=1
while c:
if a.val<b.val and c.val<b.val:res.append(cnt)
if a.val>b.val and c.val>b.val:res.append(cnt)
a,b,c=b,c,c.next
cnt+=1
n=len(res)
if n>=2:
m2=max(res)-min(res)
m1=float('inf')
for i in range(1,n):
m1=min(m1,res[i]-res[i-1])
return [m1,m2]
else:
return [-1,-1]
5916. 转化数字的最小运算数
bfs,有3个注意点
- op运算用lambda生成函数+列表存储
- deque
- vis
class Solution:
def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
op1=lambda x,y:x+y
op2=lambda x,y:x-y
op3=lambda x,y:x^y
ops=[op1,op2,op3]
q=deque([(start,0)])
vis=[0]*1001
vis[start]=1
while q:
x,step=q.popleft()
for n in nums:
for op in ops:
nx=op(x,n)
if nx==goal:
return step+1
if 0<=nx<=1000 and not vis[nx]:
vis[nx]=1
q.append((nx,step+1))
return -1
标签:周赛,return,val,int,res,self,next,力扣,265 来源: https://blog.csdn.net/weixin_45825073/article/details/122137079