Posts tagged as “string”

花花酱 LeetCode 10. Regular Expression Matching

By zxi on September 17, 2018

Problem

Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

Note:

s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like . or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

Solution 1: Recursion

Time complexity: O((|s| + |p|) * 2 ^ (|s| + |p|))

Space complexity: O(|s| + |p|)

C++

// Author: Huahua, 14 ms

class Solution {

public:

bool isMatch(string s, string p) {

return isMatch(s.c_str(), p.c_str());

}

private:

bool isMatch(const char* s, const char* p) {

if (*p == '\0') return *s == '\0';

// normal case, e.g. 'a.b','aaa', 'a'

if (p[1] != '*' || p[1] == '\0') {

// no char to match

if (*s == '\0') return false;

if (*s == *p || *p == '.')

return isMatch(s + 1, p + 1);

else

return false;

}

else {

int i = -1;

while (i == -1 || s[i] == p[0] || p[0] == '.') {

if (isMatch(s + i + 1, p + 2)) return true;

if (s[++i] == '\0') break;

}

return false;

}

return false;

}

};

花花酱 LeetCode 8. String to Integer (atoi)

By zxi on September 12, 2018

Problem

Implement atoi which converts a string to an integer.

The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned.

Note:

Only the space character ' ' is considered as whitespace character.
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2³¹, 2³¹− 1]. If the numerical value is out of the range of representable values, INT_MAX (2³¹− 1) or INT_MIN (−2³¹) is returned.

Example 1:

Input: "42"
Output: 42

Example 2:

Input: "   -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
             Then take as many numerical digits as possible, which gets 42.

Example 3:

Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.

Example 4:

Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical 
             digit or a +/- sign. Therefore no valid conversion could be performed.

Example 5:

Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
             Thefore INT_MIN (−2³¹) is returned.

Solution: Simulation

You need to handle all corner cases in order to pass…

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int myAtoi(const string& s) {

long long ans = 0;

int neg = 0;

int pos = 0;

bool p = false; // has '.'

bool a = false; // has letters

bool n = false; // has number

for (int i = 0; i < s.length(); ++i) {

if (s[i] == ' ') { if (n || neg || pos) break; }

else if (s[i] == '-') { if (n) break; else neg++; }

else if (s[i] == '+') { if (n) break; else pos++; }

else if (s[i] == '.') p = true;

else if (s[i] >= '0' && s[i] <= '9' && !p && !a) {

ans = ans * 10 + (s[i] - '0');

n = true;

} else {

a = true;

}

if (ans / 10 > INT_MAX) break; // ans can be > 64 bit

}

if (neg) ans = -ans;

if (ans > INT_MAX) ans = INT_MAX;

if (ans < INT_MIN) ans = INT_MIN;

if (pos + neg > 1) ans = 0;

return ans;

}

};

花花酱 LeetCode 6. ZigZag Conversion

By zxi on September 12, 2018

Problem

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

Example 1:
<preclass=”crayon:false”>Input: s = “PAYPALISHIRING”, numRows = 3 Output: “PAHNAPLSIIGYIR”

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I

Solution: Simulation

Store the zigzag results for each row and then output row by row.

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

string convert(string s, int nRows) {

if (nRows == 1) return s;

vector<string> ss(nRows);

int l = s.length();

int x = 0, y = 0, i = 0;

bool down = true;

while (i < l) {

ss[y] += s[i];

if (down) {

++y;

if (y == nRows) {

down = false;

y -=2;

}

} else {

--y;

if (y < 0) {

down = true;

y = 1;

}

i++;

}

string ans;

for(const string& r : ss)

ans += r;

return ans;

}

};

花花酱 LeetCode 5. Longest Palindromic Substring

By zxi on September 12, 2018

Problem

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example 2:

Input: "cbbd"
Output: "bb"

Solution: DP

Try all possible i and find the longest palindromic string whose center is i (odd case) and i / i + 1 (even case).

Time complexity: O(n^2)

Space complexity: O(1)

C++

// Author: Huahua, 16 ms, 8.7 MB

class Solution {

public:

string longestPalindrome(string s) {

const int n = s.length();

auto getLen = [&](int l, int r) {

while (l >= 0 && r < n

&& s[l] == s[r]) {

--l;

++r;

}

return r - l - 1;

};

int len = 0;

int start = 0;

for (int i = 0; i < n; ++i) {

int cur = max(getLen(i, i),

getLen(i, i + 1));

if (cur > len) {

len = cur;

start = i - (len - 1) / 2;

}

return s.substr(start, len);

}

};

Java

// Author: Huahua, 6 ms， 36.5 MB

class Solution {

public String longestPalindrome(String s) {

int len = 0;

int start = 0;

for (int i = 0; i < s.length(); ++i) {

int cur = Math.max(getLen(s, i, i),

getLen(s, i, i + 1));

if (cur > len) {

len = cur;

start = i - (cur - 1) / 2;

}

return s.substring(start, start + len);

}

private int getLen(String s, int l, int r) {

while (l >= 0 && r < s.length()

&& s.charAt(l) == s.charAt(r)) {

--l;

++r;

}

return r - l - 1;

}

Python3

# Author: Huahua, 812 ms, 12.7MB

class Solution:

def longestPalindrome(self, s: str) -> str:

n = len(s)

def getLen(l, r):

while l >= 0 and r < n and s[l] == s[r]:

l -= 1

r += 1

return r - l - 1

start = 0

length = 0

for i in range(n):

cur = max(getLen(i, i),

getLen(i, i + 1))

if cur <= length: continue

length = cur

start = i - (cur - 1) // 2

return s[start : start + length]

花花酱 LeetCode 3. Longest Substring Without Repeating Characters

By zxi on September 11, 2018

Problem

Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3.

Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

Solution: HashTable + Sliding Window

Using a hashtable to remember the last index of every char. And keep tracking the starting point of a valid substring.

start = max(start, last[s[i]] + 1)

ans = max(ans, i – start + 1)

Time complexity: O(n)

Space complexity: O(128)

C++

// Author: Huahua

class Solution {

public:

int lengthOfLongestSubstring(string s) {

const int n = s.length();

int ans = 0;

vector<int> idx(128, -1);

for (int i = 0, j = 0; j < n; ++j) {

i = max(i,idx[s[j]] + 1);

ans = max(ans, j - i + 1);

idx[s[j]] = j;

}

return ans;

}

};

Java

// Author: Huahua

class Solution {

public int lengthOfLongestSubstring(String s) {

int[] last = new int[128];

Arrays.fill(last, -1);

int start = 0;

int ans = 0;

for (int i = 0; i < s.length(); ++i) {

if (last[s.charAt(i)] != -1)

start = Math.max(start, last[s.charAt(i)] + 1);

last[s.charAt(i)] = i;

ans = Math.max(ans, i - start + 1);

}

return ans;

}

Python3

# Author: Huahua

class Solution:

def lengthOfLongestSubstring(self, s):

last = [-1] * 128

ans = 0

start = 0

for i, ch in enumerate(s):

if last[ord(ch)] != -1:

start = max(start, last[ord(ch)] + 1)

ans = max(ans, i - start + 1)

last[ord(ch)] = i

return ans