strStr
Problem Statement
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.
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)
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 your confirm with the interviewer first.
//C++ class Solution{ public: int strStr(string haystack, string needle){ if(haystack.empty() && needle.empty()) return 0; if(haystack.empty()) return -1; if(needle.empty()) return 0; if(haystack.size() < needle.size()) return -1; for(int i = 0; i < haystack.size() - needle.size() + 1; i++){ string::size_type j = 0; for(; j < needle.size(); j++){ if(haystack[i+j] != needle[j]) break; } if(j == needle.size()) return i; } return -1; } };
Two Strings Are Anagrams
Write a method anagram(s,t) to decide if two strings are anagrams or not.
Example
Given s="abcd", t="dcab", return true.
Challenge
O(n) time, O(1) extra space
//题解1 - hashmap 统计字频 class Solution{ public: bool anagram(string s, string t){ if(s.empty() || t.empty()){ return false; } if(s.size() != t.size()){ return false; } int letterCount[256] = {0}; for(int i = 0; i != s.size(); i++){ ++letterCount[s[i]]; --letterCount[t[i]]; } for(int i = 0; i != t.size(); i++){ if(letterCount[t[i]] != 0){ return false; } } return true; } };
//题解2 - 排序字符串 class Solution{ public: bool anagram(string s, string t){ if(s.empty() || t.empty()){ return false; } if(s.size() != t.size()){ return false; } sort(s.begin(), s.end()); sort(t.begin(), t.end()); if(s == t){ return true; }else{ return false; } } };
Compare Strings
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.
Example
For A = "ABCD", B = "ABC", return true.
For A = "ABCD" B = "AABC", return false.
//题目意思是问B中的所有字符是否都在A中，而不是单个字符。比如B="AABC"包含两个「A」，而A="ABCD"只包含一个「A」，故返回false. 做题时注意题意，必要时可向面试官确认。 class Solution{ public: bool compareStrings(string A, string B){ if(A.size() < B.size()){ return false; } const int AlphabetNum = 26; int letterCount[AlphabetNum] = {0}; for(int i = 0; i != A.size(); ++i){ ++letterCount[A[i] - ‘A‘]; } for(int i = 0; i != B.size(); i++){ --letterCount[B[i] - ‘A‘]; if(letterCount[B[i] - ‘A‘] < 0){ return false; } } return true; } };
Anagrams
Given an array of strings, return all groups of strings that are anagrams.
Example
Given ["lint", "intl", "inlt", "code"], return ["lint", "inlt", "intl"].
Given ["ab", "ba", "cd", "dc", "e"], return ["ab", "ba", "cd", "dc"].
Note
All inputs will be in lower-case
//题解1 - 双重for循环(TLE) class Solution{ public: vector<string> anagrams(vector<string> &strs){ if(strs.size() < 2){ return strs; } vector<string> result; vector<bool> visited(strs.size(), false); for(int s1 = 0; s1 != strs.size(); ++s1){ bool has_anagrams = false; for(int s2 = s1 + 1; s2 < strs.size(); ++s2){ if((!visited[s2]) && isAnagrams(strs[s1], strs[s2])){ result.push_back(strs[s2]); visited[s2] = true; has_anagrams = true; } } if((!visited[s1]) && has_anagrams) result.push_back(strs[s1]); } return result; } private: bool isAnagrams(string &s, string &t){ if(s.size() != t.size()){ return false; } const int AlphabetNum = 26; int letterCount[AlphabetNum] = {0}; for(int i = 0; i != s.size(); ++i){ ++letterCount[s[i] - ‘a‘]; --letterCount[t[i] - ‘a‘]; } for(int i = 0; i != t.size(); i++){ if(letterCount[t[i] - ‘a‘] < 0){ return false; } } return true; } };
//题解2 - 排序 + hashmap class Solition{ public: vector<string> anagrams(vector<string> &strs){ unordered_map<string, int> hash; for(int i = 0; i < strs.size(); i++){ string str = strs[i]; sort(str.begin(), str.end()); ++hash[str]; } vector<string> result; for(int i = 0; i < strs.size(); i++){ string str = strs[i]; sort(str.begin(), str.end()); if(hash[str] > 1){ result.push_back(strs[i]); } } return result; } };
Longest Common Substring
Given two strings, find the longest common substring.
Return the length of it.
Example
Given A="ABCD", B="CBCE", return 2.
Note
The characters in substring should occur continuously in original string.
This is different with subsequence.
class Solution{ public: int longestCommonSubstring(string &A, string &B){ if(A.empty() || B.empty()){ return 0; } int lcs = 0, lcs_temp = 0; for(int i = 0; i < A.size(); ++i){ for(int j = 0; j <B.size(); j++){ lcs_temp = 0; while((i + lcs_temp < A.size()) && (j + lcs_temp < B.size()) && (A[i + lcs_temp] == B[j + lcs_temp])){ ++lcs_temp; } if(lcs_temp > lcs){ lcs = lcs_temp; } } } return lcs; } };
Rotate String
Problem Statement
Given a string and an offset, rotate string by offset. (rotate from left to right)
Example
Given "abcdefg".
offset=0 => "abcdefg"
offset=1 => "gabcdef"
offset=2 => "fgabcde"
offset=3 => "efgabcd"
class Solution{ public: string rotateString(string A, int offset){ if(A.empty() || A.size() == 0){ return A; } int len = A.size(); offset %= len; reverse(A, 0, len - offset - 1); reverse(A, len - offset, len - 1); reverse(A, 0, len - 1); return A; } private: void reverse(string &str, int start, int end){ while(start < end){ char temp = str[start]; str[start] = str[end]; str[end] = temp; start++; end--; } } };
