也许像这样的东西会稍作调整。我自己没有运行它，但是概念应该很清楚。由于相同的空间可以被多次评估，因此也可以进行优化。
public class FindConsecutiveNumbersInGrid {public static int[][] grid = new int[][]{ {2, 5, 1, 0, 8, 0, 8}, {2, 1, 0, 9, 7, 2, 4}, {3, 3, 3, 3, 4, 6, 7}, {1, 0, 3, 4, 7, 4, 9}, {3, 3, 3, 2, 3, 1, 6}, {9, 7, 4, 1, 8, 4, 6}};public static void main(String[] args) { int maxFound = 0; int[] maxFoundPos = new int[2]; for (int i = 0; i < grid.length; i++) { for (int j = 0; j < grid[0].length; j++) { boolean[][] foundGrid = new boolean[grid.length][grid[0].length]; findConsecutive(i, j, foundGrid); int found = getFound(foundGrid); if (found > maxFound) { maxFound = found; maxFoundPos[0] = i; maxFoundPos[1] = j; } } } System.out.println(maxFoundPos[0] + " " + maxFoundPos[1]);}public static void findConsecutive(int i, int j, boolean[][] foundGrid) { foundGrid[i][j] = true; if (i < grid.length - 1 && grid[i][j] == grid[i+1][j] && !foundGrid[i+1][j]) { findConsecutive(i+1, j, foundGrid); } if (i > 0 && grid[i][j] == grid[i-1][j] && !foundGrid[i-1][j]) { findConsecutive(i-1, j, foundGrid); } if (j < grid[i].length - 1 && grid[i][j] == grid[i][j+1] && !foundGrid[i][j+1]) { findConsecutive(i, j+1, foundGrid); } if (j > 0 && grid[i][j] == grid[i][j-1] && !foundGrid[i][j-1]) { findConsecutive(i, j-1, foundGrid); }}public static int getFound(boolean[][] foundGrid) { int found = 0; for (boolean[] foundRow : foundGrid) { for (boolean foundSpace : foundRow) { if (foundSpace) found++; } } return found;}}
这将正确打印“ 2 0”。
