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 504. Base 7

By zxi on August 17, 2018

Problem

Given an integer, return its base 7 string representation.

Example 1:

Input: 100
Output: "202"

Example 2:

Input: -7
Output: "-10"

Note: The input will be in range of [-1e7, 1e7].

Solution: Simulation

Time complexity: O(logn)

Space complexity: O(logn)

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

string convertToBase7(int num) {

bool is_nagetive = num<0;

num = abs(num);

string ans;

do {

ans += ('0' + (num % 7));

num /= 7;

} while (num > 0);

if (is_nagetive)

ans += "-";

reverse(begin(ans), end(ans));

return ans;

}

};

花花酱 LeetCode 551. Student Attendance Record I

By zxi on August 17, 2018

Problem

You are given a string representing an attendance record for a student. The record only contains the following three characters:

‘A’ : Absent.
‘L’ : Late.
‘P’ : Present.

A student could be rewarded if his attendance record doesn’t contain more than one ‘A’ (absent) or more than two continuous ‘L’ (late).

You need to return whether the student could be rewarded according to his attendance record.

Example 1:

Input: "PPALLP"
Output: True

Example 2:

Input: "PPALLL"
Output: False

Solution1: Simulation

Time complexity: O(n)

Space complexity: O(1)

// Author: Huahua

// Running time: 3 ms

class Solution {

public:

bool checkRecord(string s) {

int a{0};

int l{0};

for(char c: s) {

if (c == 'A') ++a;

if (c == 'L') ++l;

else l=0;

if (a > 1 || l > 2) return false;

}

return true;

}

};

Solution 2: Regex

// Author: Huahua

// Running time: 6 ms

class Solution {

public:

bool checkRecord(string s) {

return !std::regex_search(s, std::regex("A.*A|LLL"));

}

};

花花酱 LeetCode 889. Spiral Matrix III

By zxi on August 13, 2018

Problem

On a 2 dimensional grid with R rows and C columns, we start at (r0, c0) facing east.

Here, the north-west corner of the grid is at the first row and column, and the south-east corner of the grid is at the last row and column.

Now, we walk in a clockwise spiral shape to visit every position in this grid.

Whenever we would move outside the boundary of the grid, we continue our walk outside the grid (but may return to the grid boundary later.)

Eventually, we reach all R * C spaces of the grid.

Return a list of coordinates representing the positions of the grid in the order they were visited.

Example 1:

Input: R = 1, C = 4, r0 = 0, c0 = 0
Output: [[0,0],[0,1],[0,2],[0,3]]

Example 2:

Input: R = 5, C = 6, r0 = 1, c0 = 4
Output: [[1,4],[1,5],[2,5],[2,4],[2,3],[1,3],[0,3],[0,4],[0,5],[3,5],[3,4],[3,3],[3,2],[2,2],[1,2],[0,2],[4,5],[4,4],[4,3],[4,2],[4,1],[3,1],[2,1],[1,1],[0,1],[4,0],[3,0],[2,0],[1,0],[0,0]]

Note:

1 <= R <= 100
1 <= C <= 100
0 <= r0 < R
0 <= c0 < C

Solution: Simulation

We can find out the moving sequence is ESWWNNEEESSSWWWWNNNN.

The pattern is 1,1,2,2,3,3,4,4,… steps in one direction, and turn right for 90 degrees.

directions are E,S,W,N,E,S,W,N…

Time complexity: O(max(R,C)^2)

Space complexity: O(1) or O(RC) if ans included.

// Author: Huahua

// Running time: 52 ms

class Solution {

public:

vector<vector<int>> spiralMatrixIII(int R, int C, int r0, int c0) {

vector<vector<int>> ans;

int k = 0;

int dirs[][2]{{1, 0}, {0, 1}, {-1, 0}, {0, -1}}; // ESWN

int d = 0; // E

ans.push_back({r0, c0});

if (ans.size() == R * C) return ans;

while (++k) {

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

for (int j = 0; j < k; ++j) {

c0 += dirs[d][0];

r0 += dirs[d][1];

if (c0 < 0 || c0 >= C || r0 < 0 || r0 >= R) continue;

ans.push_back({r0, c0});

if (ans.size() == R * C) return ans;

}

d = (d + 1) % 4;

}

return {};

}

};

Posts published in “Simulation”

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 504. Base 7

Problem

Solution: Simulation

花花酱 LeetCode 551. Student Attendance Record I

Problem

Solution1: Simulation

Solution 2: Regex

花花酱 LeetCode 889. Spiral Matrix III

Problem

Solution: Simulation