两个字符串是变位词
1 class Solution: 2 """ 3 @param s: The first string 4 @param b: The second string 5 @return true or false 6 """ 7 def anagram(self, s, t): 8 if len(s) == len(t): 9 d = {} 10 for i in range(len(s)): # full scan on each str 11 if s[i] in d.keys(): 12 d[s[i]] += 1 13 else: 14 d[s[i]] = 1 15 for i in range(len(s)): 16 if t[i] in d.keys(): 17 d[t[i]] -= 1 18 if d[t[i]] == 0: 19 d.pop(t[i]) 20 else: 21 return False 22 return not d 23 return False
比较字符串
1 class Solution: 2 """ 3 @param A : A string includes Upper Case letters 4 @param B : A string includes Upper Case letters 5 @return : if string A contains all of the characters in B return True else return False 6 """ 7 def compareStrings(self, A, B): 8 if len(A) >= len(B): 9 d = {} 10 for a in A: 11 if a in d: 12 d[a] += 1 13 else: 14 d[a] = 1 15 for b in B: 16 if b in d: 17 d[b] -= 1 18 if d[b] == 0: 19 d.pop(b) 20 else: 21 return False 22 return True 23 return False
字符串查找
1 class Solution: 2 def strStr(self, source, target): 3 if source != None and target != None: 4 if source == target: return 0 5 if not target: return 0 6 len_s, len_t = len(source), len(target) 7 if len_s >= len_t: 8 for i in xrange(len_s - len_t + 1): 9 if source[i] == target[0]: 10 for j in xrange(1, len_t): 11 if source[i+j] <> target[j]: 12 break 13 if j == len_t - 1: 14 return i 15 return -1
乱序字符串
First idea is brute force, keep a candidates list to store the tragets strs, then compare one by one (this compare would generate hashmap for both strs then do the compare), see solution been commented and I failed at last case for time limitation. I think over it and realized that I do not need generate mapping each time, similiar thinking like candidates list, we can keep a candidates hash mapping, so only need generate each strs hashing and compare with target mappings.
The bonus point for Python is tuple can be key as it can be hashed, so I used tuple(list) as key for the hash mapping, and I store the strs into result list while I do the checking, saved some space and traverse time. (compare the two solutions below)
Let's look back to the difference, it is typical Time or Space solution, apparently TIME goes first.
1 class Solution: 2 def anagrams(self, strs): 3 res = [] 4 if len(strs) >= 2: 5 res_map = {} 6 for s in strs: 7 count = [0] * 26 # all lower case chars 8 if s: 9 for c in s: # count chars 10 count[ord(c)-ord('a')] += 1 11 tcount = tuple(count) # tuple list as key (hash as key) 12 if tcount in res_map: 13 res_map[tuple(count)].append(s) 14 else: 15 ls = [] 16 ls.append(s) 17 res_map[tuple(count)] = ls 18 for val in res_map.values(): 19 if len(val) > 1: # anagrams 20 for s in val: 21 res.append(s) 22 return res 23 24 25 # class Solution: 26 # # @param strs: A list of strings 27 # # @return: A list of strings 28 # def anagrams(self, strs): 29 # res = [] 30 # if len(strs) >= 2: 31 # cans = [] 32 # for s in strs: 33 # flag = False 34 # if cans: 35 # for c in cans: 36 # if self.is_anagrams(s, c): 37 # if c not in res: res.append(c) 38 # res.append(s) 39 # flag = True 40 # break 41 # if not flag: 42 # cans.append(s) 43 # else: 44 # cans.append(s) 45 # return res 46 47 48 # def is_anagrams(self, s, t): 49 # if len(s) == len(t): 50 # d = {} 51 # for i in range(len(s)): # full scan on each str 52 # if s[i] in d.keys(): 53 # d[s[i]] += 1 54 # else: 55 # d[s[i]] = 1 56 # for i in range(len(s)): 57 # if t[i] in d.keys(): 58 # d[t[i]] -= 1 59 # if d[t[i]] == 0: 60 # d.pop(t[i]) 61 # else: 62 # return False 63 # return not d 64 # return False
class Solution: def anagrams(self, strs): res = [] if len(strs) >= 2: res_map = {} for s in strs: count = [0] * 26 # all lower case chars if s: # avoid "" for c in s: # count chars count[ord(c)-ord('a')] += 1 tcount = tuple(count) # tuple list as key (hash as key) if tcount in res_map: if not res_map[tcount] in res: res.append(res_map[tcount]) res.append(s) else: res_map[tcount] = s return res
最长公共子串
http://segmentfault.com/a/1190000002641054