花花酱 LeetCode 9. Palindrome Number

By zxi on September 12, 2018

Problem

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input: 121
Output: true

Example 2:

Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Follow up:

Could you solve it without converting the integer to a string?

Solution 1: Convert to string (cheating)

Time complexity: O(log10(x))

Space complexity: O(log10(x))

C++

// Author: Huahua

class Solution {

public:

bool isPalindrome(int x) {

string s = to_string(x);

return s == string(rbegin(s), rend(s));

}

};

Solution 2: Digit by Digit

Every time we compare the first and last digits of x, if they are not the same, return false. Otherwise, remove first and last digit and continue this process.

How can we achieve that via int math?

e.g. x = 9999, t = pow((10, int)log10(x)) = 1000

first digit: x / t, last digit: x % 10

then x = (x – x / t * t) / 10 removes first and last digits.

t /= 100 since we removed two digits.

x / t = 9 = 9 = x % 10, 9999 => 99

9 = 9, 99 => “”

Time complexity: O(log10(x) / 2)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

bool isPalindrome(int x) {

if (x < 0) return false;

int d = static_cast<int>(log10(x) + 1);

int t = pow(10, d - 1);

for (int i = 0; i < d / 2; ++i) {

if (x / t != x % 10) return false;

x = (x - x / t * t) / 10;

t /= 100;

}

return true;

}

};

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 7. Reverse Integer

By zxi on September 12, 2018

Problem

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2³¹, 2³¹− 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Solution: Simulation

Reverse digit by digit. Be careful about the overflow and negative numbers (especially in Python)

Time complexity: O(log(x)) ~ O(1)

Space complexity: O(log(x)) ~ O(1)

C++

// Author: Huahua

class Solution {

public:

int reverse(int x) {

int ans = 0;

while (x != 0) {

int r = x % 10;

if (ans > INT_MAX / 10 || ans < INT_MIN / 10) return 0;

ans = ans * 10 + r;

x /= 10;

}

return ans;

}

};

Java

class Solution {

public int reverse(int x) {

int ans = 0;

while (x != 0) {

int r = x % 10;

if (ans > Integer.MAX_VALUE / 10 || ans < Integer.MIN_VALUE / 10) return 0;

ans = ans * 10 + r;

x /= 10;

}

return ans;

}

Python3

class Solution:

def reverse(self, x):

min_val = -(2**31);

max_val = 2**31 - 1;

sign = -1 if x < 0 else 1

x = abs(x)

ans = 0

while x != 0:

ans = ans * 10 + (x % 10)

x //= 10

ans *= sign

if ans > max_val or ans < min_val: return 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]

Huahua's Tech Road

花花酱 LeetCode 9. Palindrome Number

Problem

Solution 1: Convert to string (cheating)

C++

Solution 2: Digit by Digit

C++

花花酱 LeetCode 8. String to Integer (atoi)

Problem

Solution: Simulation

C++

花花酱 LeetCode 7. Reverse Integer

Problem

Solution: Simulation

C++

Java

Python3

花花酱 LeetCode 6. ZigZag Conversion

Problem

Solution: Simulation

C++

花花酱 LeetCode 5. Longest Palindromic Substring

Problem

Solution: DP

C++

Java

Python3