题76(困难):
分析:
这道题其实不难,但是是我做最久的了,我居然去用res去接所有可能得值,然后再求长度导致空间暴力,我还以为是我queue的问题。。。
最后用暴力求解解的,使用双指针,移动前后指针,后指针用来找齐所有的t值,前指针用来压缩为最短值
python代码:
class Solution:def minWindow(self, s: str, t: str) -> str:#思路就是双指针,应该start一个end#加上一个map用于记录各个需要多少个,need_len用与判断还缺否if len(s)<len(t):return ''res=''need={}need_len=len(t)for c in t:need[c]=need.get(c,0)+1start,end=0,0flag=1#用于判断应该移动start还是end,1为移动end,0为移动startwhile end<len(s):while flag== 1 and end<len(s):#移动endif s[end] in need:#如果需要这个if need[s[end]]>=1:need_len-=1need[s[end]]-=1if need_len==0:flag=0#说明不需要了end+=1else:end+=1continuewhile flag == 0 and start<end:if s[start] in need:need[s[start]]+=1if need[s[start]]>0:if res=='':res=s[start:end]else:res=s[start:end] if len(res)>len(s[start:end]) else resneed_len+=1flag=1start+=1else:start+=1continuereturn res
题77(中等):
python代码:
class Solution:def combine(self, n: int, k: int) -> List[List[int]]:n_list = [i for i in range(1, n + 1)]res = []def call_back(nums, k_now, res_now):if k_now == 0:res.append(res_now)returnfor i in range(len(nums)):if i > 0 and nums[i] == nums[i - 1]:continuenums_new = nums.copy()nums_new = nums_new[i + 1:]res_new = res_now.copy()res_new.append(nums[i])call_back(nums_new, k_now - 1, res_new)call_back(n_list, k, [])return res
题78(中等):
python代码:
class Solution:def subsets(self, nums: List[int]) -> List[List[int]]:n_list=numsres = []def call_back(nums, k_now, res_now):if k_now == 0:res.append(res_now)returnfor i in range(len(nums)):if i > 0 and nums[i] == nums[i - 1]:continuenums_new = nums.copy()nums_new = nums_new[i + 1:]res_new = res_now.copy()res_new.append(nums[i])call_back(nums_new, k_now - 1, res_new)for i in range(len(nums)+1):call_back(n_list, i, [])return res
题79(中等):
python代码:
class Solution:def exist(self, board: List[List[str]], word: str) -> bool:k_len = len(word)b_row = len(board)b_col = len(board[0])notice = [[0] * b_col for i in range(b_row)]def call_back(notice, x, y, k): # x为当前的横坐标,y为当前纵坐标,k为word的第几个if board[x][y] != word[k]:return Falseif k == k_len-1:return Truenotice[x][y] = 1if x - 1 >= 0 and notice[x - 1][y] == 0:if call_back(notice, x - 1, y, k + 1):return Trueif x + 1 < b_row and notice[x + 1][y] == 0:if call_back(notice, x + 1, y, k + 1):return Trueif y - 1 >= 0 and notice[x][y - 1] == 0:if call_back(notice, x, y - 1, k + 1):return Trueif y + 1 < b_col and notice[x][y + 1] == 0:if call_back(notice, x, y + 1, k + 1):return Truenotice[x][y]=0return Falsefor i in range(b_row):for j in range(b_col):if call_back(notice, i, j, 0):return Truereturn False
题80(中等):
python代码:
class Solution:def removeDuplicates(self, nums: List[int]) -> int:# 还记得当初有个3国旗问题吗,那个前两指针在开头的---->T75p0, p1 = 0, 0cout, num = 0, 0while p1 < len(nums):if nums[p1] == num:# 表示这个数要放后面if cout >= 2:p1 = p1 + 1else:# 表示这个数不放后面nums[p0], nums[p1] = nums[p1], nums[p0]p0 += 1p1 += 1cout += 1else:# 表示这个数是新的,不放后面cout = 1num = nums[p1]nums[p0], nums[p1] = nums[p1], nums[p0]p0 += 1p1+=1return p0