Huahua's Tech Road

花花酱 LeetCode 46. Permutations

By zxi on October 2, 2019

Given a collection of distinct integers, return all possible permutations.

Example:

Input: [1,2,3]
Output:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

Solution: DFS

Time complexity: O(n!)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

vector<vector<int>> permute(vector<int>& nums) {

const int n = nums.size();

vector<vector<int>> ans;

vector<int> used(n);

vector<int> path;

function<void(int)> dfs = [&](int d) {

if (d == n) {

ans.push_back(path);

return;

}

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

if (used[i]) continue;

used[i] = 1;

path.push_back(nums[i]);

dfs(d + 1);

path.pop_back();

used[i] = 0;

}

};

dfs(0);

return ans;

}

};

花花酱 LeetCode 48. Rotate Image

By zxi on October 2, 2019

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Note:

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

Example 2:

Given input matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

Solution: 2 Passes

First pass: mirror around diagonal
Second pass: mirror around y axis

Time complexity: O(n^2)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

void rotate(vector<vector<int>>& matrix) {

const int n = matrix.size();

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

for (int j = i + 1; j < n; ++j)

swap(matrix[i][j], matrix[j][i]);

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

reverse(begin(matrix[i]), end(matrix[i]));

}

};

花花酱 LeetCode 49. Group Anagrams

By zxi on October 2, 2019

Given an array of strings, group anagrams together.

Example:

Input: ["eat", "tea", "tan", "ate", "nat", "bat"],
Output:
[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]

Note:

All inputs will be in lowercase.
The order of your output does not matter.

Solution: HashTable

The sorted word will be the key of each group

Time complexity: O(sum(l*log(l)))
Space complexity: O(sum(l))

C++

// Author: Huahua

class Solution {

public:

vector<vector<string>> groupAnagrams(vector<string>& strs) {

vector<vector<string>> ans;

unordered_map<string, vector<int>> m;

for (int i = 0; i < strs.size(); ++i) {

string c = strs[i];

sort(begin(c), end(c));

m[c].push_back(i);

}

for (const auto& kv : m) {

ans.push_back({});

for (int i : kv.second)

ans.back().push_back(strs[i]);

}

return ans;

}

};

花花酱 LeetCode 52. N-Queens II

By zxi on October 2, 2019

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

Given an integer n, return the number of distinct solutions to the n-queens puzzle.

Example:

Input: 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown below.
[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

Solution: DFS

Time complexity: O(n!)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int totalNQueens(int n) {

vector<int> cols(n);

vector<int> diag1(2 * n - 1);

vector<int> diag2(2 * n - 1);

int ans = 0;

function<void(int)> dfs = [&](int r) {

if (r == n) {

++ans;

return;

}

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

int& c = cols[i];

int& d1 = diag1[r + i];

int& d2 = diag2[r - i + n - 1];

if (c || d1 || d2) continue;

c = d1 = d2 = 1;

dfs(r + 1);

c = d1 = d2 = 0;

}

};

dfs(0);

return ans;

}

};

花花酱 LeetCode 68. Text Justification

By zxi on October 1, 2019

Given an array of words and a width maxWidth, format the text such that each line has exactly maxWidth characters and is fully (left and right) justified.

You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ' ' when necessary so that each line has exactly maxWidth characters.

Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line do not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right.

For the last line of text, it should be left justified and no extra space is inserted between words.

Note:

A word is defined as a character sequence consisting of non-space characters only.
Each word’s length is guaranteed to be greater than 0 and not exceed maxWidth.
The input array words contains at least one word.

Example 1:

Input:
words = ["This", "is", "an", "example", "of", "text", "justification."]
maxWidth = 16
Output:
[
   "This    is    an",
   "example  of text",
   "justification.  "
]

Example 2:

Input:
words = ["What","must","be","acknowledgment","shall","be"]
maxWidth = 16
Output:
[
  "What   must   be",
  "acknowledgment  ",
  "shall be        "
]
Explanation: Note that the last line is "shall be    " instead of "shall     be",
             because the last line must be left-justified instead of fully-justified.
             Note that the second line is also left-justified becase it contains only one word.

Example 3:

Input:
words = ["Science","is","what","we","understand","well","enough","to","explain",
         "to","a","computer.","Art","is","everything","else","we","do"]
maxWidth = 20
Output:
[
  "Science  is  what we",
  "understand      well",
  "enough to explain to",
  "a  computer.  Art is",
  "everything  else  we",
  "do                  "
]

Solution: Simulation

Time complexity: O(sum(len(s))
Space complexity: O(sum(len(s))

C++

// Author: Huahua

class Solution {

public:

vector<string> fullJustify(vector<string> &words, int L) {

size_t i = 0, n = words.size();

vector<string> ans;

while (i < n) {

bool last_line = false;

int ll = 0;

vector<string_view> tw;

while (ll <= L && i < n) {

size_t wl = words[i].length();

int tll = ll + (tw.size() ? 1 : 0) + wl;

if (tll > L) break;

ll = tll;

tw.push_back(string_view(words[i]));

if (++i == n) last_line = true;

if (ll == L) break;

}

string line;

if (!last_line) {

int tl = 0;

for(const auto& w : tw) tl += w.length();

int avg_space = tw.size()==1 ? 0 : ((L-tl) / (tw.size() - 1));

int extra_space = tw.size()==1 ? 0 : (L - avg_space * (tw.size() - 1) - tl);

for (const auto& w : tw) {

if (line.length() > 0) {

line.append(avg_space, ' ');

if (extra_space > 0) {

line += ' ';

--extra_space;

}

line += w;

}

} else {

for (const auto& w : tw) {

if (line.length() > 0)

line += ' ';

line += w;

}

line.append(L - line.length(), ' ');

ans.push_back(std::move(line));

}

return ans;

}

};

花花酱 LeetCode 46. Permutations

Solution: DFS

C++

Related Problems

花花酱 LeetCode 48. Rotate Image

Solution: 2 Passes

C++

花花酱 LeetCode 49. Group Anagrams

Solution: HashTable

C++

花花酱 LeetCode 52. N-Queens II

Solution: DFS

C++

Related Problems

花花酱 LeetCode 68. Text Justification

C++