Problem

According to the Wikipedia’s article: “The Game of Life, also known simply as Life, is a cellular automaton devised by the British mathematician John Horton Conway in 1970.”

Given a board with m by n cells, each cell has an initial state live (1) or dead (0). Each cell interacts with its eight neighbors (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article):

Any live cell with fewer than two live neighbors dies, as if caused by under-population.

Any live cell with two or three live neighbors lives on to the next generation.

Any live cell with more than three live neighbors dies, as if by over-population..

Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.

Write a function to compute the next state (after one update) of the board given its current state. The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously.

Example:

Input: [ [0,1,0], [0,0,1], [1,1,1], [0,0,0] ] Output: [ [0,0,0], [1,0,1], [0,1,1], [0,1,0] ]

Follow up:

Could you solve it in-place? Remember that the board needs to be updated at the same time: You cannot update some cells first and then use their updated values to update other cells.

In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches the border of the array. How would you address these problems?

Solution: Simulation

Time complexity: O(mn)

Space complexity: O(1)

// Author: Huahua

// Running time: 5 ms

class Solution {

public:

void gameOfLife(vector<vector<int>>& board) {

int m = board.size();

int n = m ? board[0].size() : 0;

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

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

int lives = 0;

// Scan the 3x3 region including (j, i).

for (int y = max(0, i - 1); y < min(m, i + 2); ++y)

for (int x = max(0, j - 1); x < min(n, j + 2); ++x)

lives += board[y][x] & 1;

if (lives == 3 || lives - board[i][j] == 3) board[i][j] |= 0b10;

}

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

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

board[i][j] >>= 1;

}

};

Problem

题目大意：对一个string进行in-place的run length encoding。

https://leetcode.com/problems/string-compression/description/

Given an array of characters, compress it in-place.

The length after compression must always be smaller than or equal to the original array.

Every element of the array should be a character (not int) of length 1.

After you are done modifying the input array in-place, return the new length of the array.

Follow up:
Could you solve it using only O(1) extra space?

Example 1:

Input:
["a","a","b","b","c","c","c"]

Output:
Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]

Explanation:
"aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".

Example 2:

Input:
["a"]

Output:
Return 1, and the first 1 characters of the input array should be: ["a"]

Explanation:
Nothing is replaced.

Example 3:

Input:
["a","b","b","b","b","b","b","b","b","b","b","b","b"]

Output:
Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"].

Explanation:
Since the character "a" does not repeat, it is not compressed. "bbbbbbbbbbbb" is replaced by "b12".
Notice each digit has it's own entry in the array.

Note:

All characters have an ASCII value in [35, 126].
1 <= len(chars) <= 1000.

Solution

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 9 ms

class Solution {

public:

int compress(vector<char>& chars) {

const int n = chars.size();

int p = 0;

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

int count = 1;

while (i < n && chars[i] == chars[i - 1]) { ++i; ++count; }

chars[p++] = chars[i - 1];

if (count == 1) continue;

for (char c : to_string(count))

chars[p++] = c;

}

return p;

}

};

Posts tagged as “in place”

花花酱 LeetCode 289. Game of Life

Problem

Solution: Simulation

花花酱 LeetCode 443. String Compression

Problem

Solution