723 Candy Crush

723. Candy Crush

1. Question

This question is about implementing a basic elimination algorithm for Candy Crush.
Given a 2D integer arrayboardrepresenting the grid of candy, different positive integersboard[i][j]represent different types of candies. A value ofboard[i][j] = 0represents 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 astable stateby crushing candies according to the following rules:
  1. 1.
    If three or more candies of the same type are adjacent vertically or horizontally, "crush" them all at the same time - these positions become empty.
  2. 2.
    After crushing all candies simultaneously, if an empty space on the board has candies on top of itself, then these candies will drop until they hit a candy or bottom at the same time. (No new candies will drop outside the top boundary.)
  3. 3.
    After the above steps, there may exist more candies that can be crushed. If so, you need to repeat the above steps.
  4. 4.
    If there does not exist more candies that can be crushed (ie. the board is stable), then return the current board.
You need to perform the above rules until the board becomes stable, then return the current board.
Example 1:
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Input:
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board =
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[[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]]
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Output:
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[[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]]
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Explanation:
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Note:
  1. 1.
    The length ofboardwill be in the range [3, 50].
  2. 2.
    The length ofboard[i]will be in the range [3, 50].
  3. 3.
    Eachboard[i][j]will initially start as an integer in the range [1, 2000].

2. Implementation

(1) Two Pointers
思路: Two Pointers的思想指体现在Candy Drop的时候,对每一列,将非零的数往下drop,而将0往上移。
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class Solution {
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public int[][] candyCrush(int[][] board) {
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if (board == null || board.length == 0) {
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return new int[0][0];
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}
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int m = board.length, n = board[0].length;
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while (true) {
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Set<Coordinate> set = new HashSet<>();
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// Find index where candy can be crushed
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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if (board[i][j] == 0) continue;
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int top = i, bottom = i, left = j, right = j;
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while (top >= 0 && top > i - 3 && board[top][j] == board[i][j]) --top;
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while (bottom < m && bottom < i + 3 && board[bottom][j] == board[i][j]) ++bottom;
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while (left >= 0 && left > j - 3 && board[i][left] == board[i][j]) --left;
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while (right < n && right < j + 3 && board[i][right] == board[i][j]) ++right;
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if (right - left > 3 || bottom - top > 3) {
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set.add(new Coordinate(i, j));
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}
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}
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}
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// Board is in stable state
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if (set.isEmpty()) break;
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// Mark index as 0
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for (Coordinate c : set) {
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board[c.x][c.y] = 0;
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}
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// Use Two Pointers to move zero to top, and keep number in order in column
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for (int col = 0; col < n; col++) {
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int index = m - 1;
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for (int row = m - 1; row >= 0; row--) {
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if (board[row][col] != 0) {
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swap(board, index, row, col);
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--index;
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}
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}
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}
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}
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return board;
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}
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public void swap(int[][] board, int row1, int row2, int col) {
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if (row1 == row2) return;
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int temp = board[row1][col];
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board[row1][col] = board[row2][col];
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board[row2][col] = temp;
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}
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class Coordinate {
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int x;
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int y;
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public Coordinate(int x, int y) {
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this.x = x;
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this.y = y;
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}
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}
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}
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3. Time & Space Complexity

时间复杂度O(mn), 空间复杂度O(mn)