标签:leetcode mes only array 题解 ISE inpu https bsp
原题链接在这里:https://leetcode.com/problems/candy-crush/
题目:
This question is about implementing a basic elimination algorithm for Candy Crush.
Given a 2D integer array board
representing the grid of candy, different positive integers board[i][j]
represent different types of candies. A value of board[i][j] = 0
represents that the cell at position (i, j)
is empty. The given board represents the state of the game following the player‘s move. Now, you need to restore the board to a stable state by crushing candies according to the following rules:
You need to perform the above rules until the board becomes stable, then return the current board.
Example:
Input: board = [[110,5,112,113,114],[210,211,5,213,214],[310,311,3,313,314],[410,411,412,5,414],[5,1,512,3,3],[610,4,1,613,614],[710,1,2,713,714],[810,1,2,1,1],[1,1,2,2,2],[4,1,4,4,1014]] Output: [[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[110,0,0,0,114],[210,0,0,0,214],[310,0,0,113,314],[410,0,0,213,414],[610,211,112,313,614],[710,311,412,613,714],[810,411,512,713,1014]] Explanation:
Note:
board
will be in the range [3, 50].board[i]
will be in the range [3, 50].board[i][j]
will initially start as an integer in the range [1, 2000].题解:
There are two steps.
Step 1: Mark 3 adjacent candies. Check if there are 3 adjacent candies. First check row by row, then column by column.
If there are, mark these values as negative.
Step 2: Crush them. Rewrite board with only positive numbers.
If there is crushing, that means there may be another round of crash, use recursion. Otherwise, there wouldn‘t be another round of crash, return the board.
Time Comlexity: O(M^2*n^2). m = board.length. n = board[0].length.
Each crash, there would be 3 crashed at minimum. Totally there are m*n candies. So recursion could run for m*n/3 times.
Each recursion, it takes O(m*n).
Space: O(1).
AC Java:
1 class Solution { 2 public int[][] candyCrush(int[][] board) { 3 if(board == null || board.length == 0 | board[0].length == 0){ 4 return board; 5 } 6 7 boolean todo = false; 8 int m = board.length; 9 int n = board[0].length; 10 for(int i = 0; i<m; i++){ 11 for(int j = 0; j<n-2; j++){ 12 int val = Math.abs(board[i][j]); 13 if(val!=0 && val==Math.abs(board[i][j+1]) && val==Math.abs(board[i][j+2])){ 14 todo = true; 15 board[i][j] = board[i][j+1] = board[i][j+2] = -val; 16 } 17 } 18 } 19 20 for(int j = 0; j<n; j++){ 21 for(int i = 0; i<m-2; i++){ 22 int val = Math.abs(board[i][j]); 23 if(val!=0 && val==Math.abs(board[i+1][j]) && val==Math.abs(board[i+2][j])){ 24 todo = true; 25 board[i][j] = board[i+1][j] = board[i+2][j] = -val; 26 } 27 } 28 } 29 30 for(int j = 0; j<n; j++){ 31 int br = m-1; 32 for(int i = m-1; i>=0; i--){ 33 if(board[i][j] > 0){ 34 board[br--][j] = board[i][j]; 35 } 36 } 37 38 while(br>=0){ 39 board[br--][j] = 0; 40 } 41 } 42 43 return todo ? candyCrush(board) : board; 44 } 45 }
标签:leetcode mes only array 题解 ISE inpu https bsp
原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11421584.html