關(guān)于我的 Leetcode 題目解答喝峦，代碼前往 Github：https://github.com/chenxiangcyr/leetcode-answers
LeetCode題目：661 Image Smoother
Given a 2D integer matrix M representing the gray scale of an image, you need to design a smoother to make the gray scale of each cell becomes the average gray scale (rounding down) of all the 8 surrounding cells and itself. If a cell has less than 8 surrounding cells, then use as many as you can.
Example 1:
Input:
[[1,1,1],
[1,0,1],
[1,1,1]]
Output:
[[0, 0, 0],
[0, 0, 0],
[0, 0, 0]]
Explanation:
For the point (0,0), (0,2), (2,0), (2,2): floor(3/4) = floor(0.75) = 0
For the point (0,1), (1,0), (1,2), (2,1): floor(5/6) = floor(0.83333333) = 0
For the point (1,1): floor(8/9) = floor(0.88888889) = 0
Note:
- The value in the given matrix is in the range of [0, 255].
- The length and width of the given matrix are in the range of [1, 150].
分析：
題中看出數(shù)字的范圍是[0, 255]咱揍，一個整型是32位部宿，255只占用后8位。我們可以利用中間的8位來存儲計算結(jié)果，即00000000-00000000-計算結(jié)果-原始數(shù)據(jù)
對于一個數(shù)字A棉胀，獲得后8位，可以通過A&0xFF
對于一個數(shù)字A冀膝，將某個數(shù)字B存儲到A中間的8位唁奢，可以通過A = A | (B<<8)
完整代碼如下：
class Solution {
public int[][] imageSmoother(int[][] M) {
int rows = M.length;
int columns = M[0].length;
for(int i = 0; i < rows; i++) {
for(int j = 0; j < columns; j++) {
// 對于一個數(shù)字A，獲得后8位窝剖，可以通過A&0xFF
int sum = M[i][j]&0xFF;
int count = 1;
// (i-1, j-1)
if(i > 0 && j > 0) {
sum = sum + (M[i-1][j-1]&0xFF);
count++;
}
// (i-1, j)
if(i > 0) {
sum = sum + (M[i-1][j]&0xFF);
count++;
}
// (i-1, j+1)
if(i > 0 && j < columns - 1) {
sum = sum + (M[i-1][j+1]&0xFF);
count++;
}
// (i, j - 1)
if(j > 0) {
sum = sum + (M[i][j-1]&0xFF);
count++;
}
// (i, j + 1)
if(j < columns - 1) {
sum = sum + (M[i][j+1]&0xFF);
count++;
}
// (i+1, j-1)
if(i < rows - 1 && j > 0) {
sum = sum + (M[i+1][j-1]&0xFF);
count++;
}
// (i+1, j)
if(i < rows - 1) {
sum = sum + (M[i+1][j]&0xFF);
count++;
}
// (i+1, j+1)
if(i < rows - 1 && j < columns - 1) {
sum = sum + (M[i+1][j+1]&0xFF);
count++;
}
// 對于一個數(shù)字A麻掸，將某個數(shù)字B存儲到A中間的8位，可以通過A = A | (B<<8)
M[i][j] |= ((sum / count) << 8);
}
}
for(int i = 0; i < rows; i++) {
for(int j = 0; j < columns; j++) {
M[i][j] >>= 8;
}
}
return M;
}
}
LeetCode題目：289. Game of Life
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.
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?
分析：
To solve it in place, we use 2 bits to store 2 states:
[2nd bit, 1st bit] = [next state, current state]
- 00 dead (next) <- dead (current)
- 01 dead (next) <- live (current)
- 10 live (next) <- dead (current)
- 11 live (next) <- live (current)
In the beginning, every cell is either 00 or 01.
Notice that 1st state is independent of 2nd state.
Imagine all cells are instantly changing from the 1st to the 2nd state, at the same time.
Let’s count # of neighbors from 1st state and set 2nd state bit.
Since every 2nd state is by default dead, no need to consider transition 01 -> 00.
In the end, delete every cell’s 1st state by doing >> 1.
For each cell’s 1st bit, check the 8 pixels around itself, and set the cell’s 2nd bit.
Transition 01 -> 11: when board == 1 and lives >= 2 && lives <= 3.
Transition 00 -> 10: when board == 0 and lives == 3.
To get the current state, simply do board[i][j] & 1
To get the next state, simply do board[i][j] >> 1
完整代碼如下：
public void gameOfLife(int[][] board) {
if (board == null || board.length == 0) return;
int m = board.length, n = board[0].length;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int lives = liveNeighbors(board, m, n, i, j);
// In the beginning, every 2nd bit is 0;
// So we only need to care about when will the 2nd bit become 1.
if (board[i][j] == 1 && lives >= 2 && lives <= 3) {
board[i][j] = 3; // Make the 2nd bit 1: 01 ---> 11
}
if (board[i][j] == 0 && lives == 3) {
board[i][j] = 2; // Make the 2nd bit 1: 00 ---> 10
}
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
board[i][j] >>= 1; // Get the 2nd state.
}
}
}
public int liveNeighbors(int[][] board, int m, int n, int i, int j) {
int lives = 0;
for (int x = Math.max(i - 1, 0); x <= Math.min(i + 1, m - 1); x++) {
for (int y = Math.max(j - 1, 0); y <= Math.min(j + 1, n - 1); y++) {
lives += board[x][y] & 1;
}
}
lives -= board[i][j] & 1;
return lives;
}
