[Lintcode Ladder2] String
1. 158. Valid Anagram
Description
Write a method anagram(s,t) to decide if two strings are anagrams or not.
Have you met this question in a real interview?
Clarification
What is Anagram?
Two strings are anagram if they can be the same after change the order of characters.
Example
Given s = “abcd”, t = “dcab”, return true.
Given s = “ab”, t = “ab”, return true.
Given s = “ab”, t = “ac”, return false.
Challenge
O(n) time, O(1) extra space
我的代码：
class Solution:
def getdic(self, s):
d_s = {}
for key in s:
d_s[key] = d_s.get(key,0)+1
return d_s
def isAnagram(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
return self.getdic(s) == self.getdic(t)
其他人的代码：
def isAnagram3(self, s, t):
return sorted(s) == sorted(t)
注意：此时得到的是一个数组
思路：
我想的是，如果顺序不重要的话，就是相应的字母都有，而且数量一样。所以可以用dictionary来实现（数组亦可）。看到一行代码的思路是顺序不重要的话，可以按照相同的规则重新排序，最后结果完全一样就可以了。
2. 55. Compare Strings
Description
Compare two strings A and B, determine whether A contains all of the characters in B.
The characters in string A and B are all Upper Case letters.
The characters of B in A are not necessary continuous or ordered.
Have you met this question in a real interview?
Example
For A = “ABCD”, B = “ACD”, return true.
For A = “ABCD”, B = “AABC”, return false.
Related Problems
我的代码：
def compareStrings(self, A, B):
# write your code here
dic_A = {}
for key in A:
dic_A[key] = dic_A.get(key,0)+1
dic_B = {}
for key in B:
dic_B[key] = dic_B.get(key,0)+1
for keyb in B:
if dic_A.get(keyb) == None:
return False
else:
if dic_A.get(keyb) < dic_B[keyb]:
return False
return True
思路：
大抵同上一题
3.Implement strStr()
Description
For a given source string and a target string, you should output the first index(from 0) of target string in source string.
If target does not exist in source, just return -1.
Have you met this question in a real interview?
Clarification
Do I need to implement KMP Algorithm in a real interview?
Not necessary. When you meet this problem in a real interview, the interviewer may just want to test your basic implementation ability. But make sure you confirm with the interviewer first.
Example
If source = “source” and target = “target”, return -1.
If source = “abcdabcdefg” and target = “bcd”, return 1.
Challenge
O(n2) is acceptable. Can you implement an O(n) algorithm? (hint: KMP)
我的代码
暴力求解
def strStr(self, source, target):
# Write your code here
h = 0
t = 0
l_s = len(source)
l_t = len(target)
if source == target:
return 0
for i in range(l_s-l_t+1):
if source[i:i+l_t] == target:
return i
return -1
学习 KMP
参考文章：从头到尾彻底理解KMP（2014年8月22日版）
该算法的重点是，构建一个部分匹配表，使得无法完整匹配时，不是单步移动，而是移动最长的前缀后缀匹配位置。
参考代码
#KMP
# s 原字符串
# p 子串
def kmp_match(s, p):
m = len(s); n = len(p)
cur = 0#起始指针cur
table = partial_table(p)
while cur<=m-n:
for i in range(n):
if s[i+cur]!=p[i]:
cur += max(i - table[i-1], 1)#有了部分匹配表,我们不只是单纯的1位1位往右移,可以一次移动多位
break
else:
return True
return False
#部分匹配表
def partial_table(p):
'''partial_table("ABCDABD") -> [0, 0, 0, 0, 1, 2, 0]'''
prefix = set()
postfix = set()
ret = [0]
for i in range(1,len(p)):
prefix.add(p[:i])
postfix = {p[j:i+1] for j in range(1,i+1)}
ret.append(len((prefix&postfix or {''}).pop()))
return ret
print kmp_match("BBC ABCDAB ABCDABCDABDE", "ABCDABD")
4. 171. Anagrams
Description
Given an array of strings, return all groups of strings that are anagrams.
All inputs will be in lower-case
Have you met this question in a real interview?
Example
Given [“lint”, “intl”, “inlt”, “code”], return [“lint”, “inlt”, “intl”].
Given [“ab”, “ba”, “cd”, “dc”, “e”], return [“ab”, “ba”, “cd”, “dc”].
Challenge
What is Anagram?
Two strings are anagram if they can be the same after change the order of characters.
Related Problems
我的代码：
class Solution:
"""
@param strs: A list of strings
@return: A list of strings
"""
def anagrams(self, strs):
# write your code here
dic = {}
#将字符按照字母表顺序，重新排序，作为它的anagram特征
for item in strs:
item_ = ''
for i in sorted(item):
item_ += i
dic[item_] = dic.get(item_, 0) + 1
res = []
for item in strs:
item_ = ''
for i in sorted(item):
item_ += i
#若特征相同，则为anagram
if dic[item_] > 1:
res.append(item)
return res
78. Longest Common Prefix
Description
Given k strings, find the longest common prefix (LCP).
Example
For strings “ABCD”, “ABEF” and “ACEF”, the LCP is “A”
For strings “ABCDEFG”, “ABCEFG” and “ABCEFA”, the LCP is “ABC”
我的代码
class Solution:
"""
@param strs: A list of strings
@return: The longest common prefix
"""
def longestCommonPrefix(self, strs):
if strs == []:
return ""
l = max([len(i) for i in strs])
res = ""
for i in range(0,l):
if len(set([prefix[:i+1] for prefix in strs])) == 1:
res = strs[0][:i+1]
return res
出错
- 没有考虑strs == [] 的情况
- 一开始没考虑到串长为1的情况-> 边界考虑不足
