记录了初步解题思路 以及本地实现代码；并不一定为最优 也希望大家能一起探讨 一起进步
目录
12/11 1631. 最小体力消耗路径
dijstra
最小堆 找到当前最小的路径
import heapq
def minimumEffortPath(heights):
"""
:type heights: List[List[int]]
:rtype: int
"""
m,n = len(heights),len(heights[0])
dist = [0]+[float("inf")]*(m*n-1)
q = [(0,0,0)]
path =set()
while q:
d,x,y = heapq.heappop(q)
idx = x*n+y
if idx in path:
continue
if (x,y)==(m-1,n-1):
break
path.add(idx)
for xx,yy in [(x-1,y),(x+1,y),(x,y-1),(x,y+1)]:
if 0<=xx<m and 0<=yy<n:
maxd = max(d,abs(heights[x][y]-heights[xx][yy]))
if maxd<=dist[xx*n+yy]:
dist[xx*n+yy] = maxd
heapq.heappush(q,(dist[xx*n+yy],xx,yy))
return dist[m*n-1]
12/12 2454. 下一个更大元素 IV
将数值从大到小排序 并记录其坐标(num,i)
使用一个有序队列s记录已经处理过的坐标
对于当前处理的nums[x] s中所有数值都比其大 只要找到比x大的下下个元素值
即为nums[x]对应的第二大元素坐标
def secondGreaterElement(nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
from sortedcontainers import SortedList
l = [(x,i) for i,x in enumerate(nums)]
l.sort(key=lambda x:-x[0])
s = SortedList()
n = len(nums)
ans = [-1]*n
for _,i in l:
loc = s.bisect_right(i)
if loc+1<len(s):
ans[i] = nums[s[loc+1]]
s.add(i)
return ans
12/13 2697. 字典序最小回文串
左右指针依次比较 对应的位置需要相同且使用字典序小的字母
def makeSmallestPalindrome(s):
"""
:type s: str
:rtype: str
"""
l,r=0,len(s)-1
li = [""]*len(s)
while l<=r:
c = s[l]
if s[r]<s[l]:
c=s[r]
li[l]=li[r]=c
l+=1
r-=1
return ''.join(li)
12/14 2132. 用邮票贴满网格图
二维前缀和判断是否可以贴邮票
二维差分判断是否所有空格子被覆盖
def possibleToStamp(grid, stampHeight, stampWidth):
"""
:type grid: List[List[int]]
:type stampHeight: int
:type stampWidth: int
:rtype: bool
"""
m,n=len(grid),len(grid[0])
psum = [[0]*(n+2) for _ in range(m+2)]
diff = [[0]*(n+2) for _ in range(m+2)]
for i in range(1,m+1):
for j in range(1,n+1):
psum[i][j] = psum[i-1][j]+psum[i][j-1]-psum[i-1][j-1]+grid[i-1][j-1]
for i in range(1,m+2-stampHeight):
for j in range(1,n+2-stampWidth):
x = i+stampHeight-1
y = j+stampWidth-1
if psum[x][y]-psum[x][j-1]-psum[i-1][y]+psum[i-1][j-1]==0:
diff[i][j]+=1
diff[i][y+1]-=1
diff[x+1][j]-=1
diff[x+1][y+1]+=1
for i in range(1,m+1):
for j in range(1,n+1):
diff[i][j] += diff[i-1][j]+diff[i][j-1]-diff[i-1][j-1]
if diff[i][j]==0 and grid[i-1][j-1]==0:
return False
return True
12/15 2415. 反转二叉树的奇数层
BFS
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def reverseOddLevels(root):
"""
:type root: Optional[TreeNode]
:rtype: Optional[TreeNode]
"""
li = [root]
lev = 0
while len(li)>0:
if lev%2==1:
l,r=0,len(li)-1
while l<r:
x,y = li[l],li[r]
x.val,y.val=y.val,x.val
l+=1
r-=1
tmp=[]
for node in li:
if not node.left:
break
tmp.append(node.left)
tmp.append(node.right)
lev+=1
li = tmp
return root
12/16 2276. 统计区间中的整数数目
线段树
l,r为左右端点 cnt为区间内整数个数
如果当前区间全覆盖 则cnt=r-l+1
class CountIntervals(object):
def __init__(self,l=1,r=10**9):
self.left = self.right = None
self.l,self.r,self.cnt=l,r,0
def add(self, left, right):
"""
:type left: int
:type right: int
:rtype: None
"""
if self.cnt == self.r-self.l+1:
return
if left<=self.l and self.r<=right:
self.cnt = self.r-self.l+1
return
mid = (self.l+self.r)//2
if self.left is None:
self.left = CountIntervals(self.l,mid)
if self.right is None:
self.right = CountIntervals(mid+1,self.r)
if left<=mid:
self.left.add(left,right)
if mid<right:
self.right.add(left,right)
self.cnt = self.left.cnt+self.right.cnt
def count(self):
"""
:rtype: int
"""
return self.cnt
12/17 746. 使用最小花费爬楼梯
当前i楼梯可以由 i-1,i-2到达
def minCostClimbingStairs(cost):
"""
:type cost: List[int]
:rtype: int
"""
f1,f2=0,0
for i in cost:
f1,f2=i+min(f1,f2),f1
ret = min(f1,f2)
return ret
