给定一个矩阵,寻找连通域个数:前后左右相同为连通
ex:
0 1 0 1
0 1 1 1
0 0 1 0
0 1 0 0
输出2
利用深度搜索思路:
1 public static int getCount(int[][] A) { 2 int result = 0; 3 for (int i = 0; i < A.length; i++) { 4 for (int j = 0; j < A[0].length; j++) { 5 if (A[i][j] == 1) { 6 result++; 7 erase(A, i, j); 8 } 9 } 10 } 11 return result; 12 } 13 14 public static void erase(int[][] A, int i, int j) { 15 A[i][j] = 0; 16 while (i - 1 >= 0 && A[i - 1][j] == 1) { 17 erase(A, i - 1, j); 18 } 19 while (i + 1 < A.length && A[i + 1][j] == 1) { 20 erase(A, i + 1, j); 21 } 22 while (j - 1 >= 0 && A[i][j - 1] == 1) { 23 erase(A, i, j - 1); 24 } 25 while (j + 1 < A[0].length && A[i][j + 1] == 1) { 26 erase(A, i, j + 1); 27 } 28 29 }