Number of Islands 岛屿个数
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给一个01矩阵,求不同的岛屿的个数。
0代表海,1代表岛,如果两个1相邻,那么这两个1属于同一个岛。我们只考虑上下左右为相邻。
样例在矩阵:
[ [1, 1, 0, 0, 0], [0, 1, 0, 0, 1], [0, 0, 1, 1, 1], [0, 0, 0, 0, 0], [0, 0, 0, 0, 1]]
中有 3 个岛,两片湖。
复杂度:
时间 O(NM) 空间 O(max(N,M)) 递归栈空间
编程思想:
遍历矩阵的每一个点,对每个点都尝试进行一次深度优先搜索,如果搜索到1,就继续向它的四周搜索。同时我们每找到一个1,就将其标为0,这样就能把整个岛屿变成0。我们只要记录对矩阵遍历时能进入多少次搜索,就代表有多少个岛屿。
代码如下:
public class NumOfIsland {public static void main(String[] args) {char[][] grid = {{'1', '1', '0', '0', '0'},{'0', '1', '0', '0', '1'},{'0', '1', '1', '1', '1'},{'0', '0', '0', '0', '0'},{'0', '0', '0', '0', '1'}};System.out.println(NumOfIsland.numIsland(grid));//System.out.println(NumOfIsland.numOfSea(grid));}public static int numIsland(char[][] grid) {int num = 0;if (grid.length == 0) {return num;}for (int i = 0; i < grid.length; i++) {for (int j = 0; j < grid[i].length; j++) {if (grid[i][j] == '1') {grid = changeO(grid, i, j);num++;}}}return num;}public static int numOfSea(char[][] grid) {int num = 0;if (grid.length == 0) {return num;}for (int i = 0; i < grid.length; i++) {for (int j = 0; j < grid[i].length; j++) {if (grid[i][j] == '0') {grid = changeZ(grid, i, j);num++;}}}return num;}public static char[][] changeO(char[][] grid,int i,int j) {grid[i][j] = 0;//修改为0if (i > 0 && grid[i-1][j] == '1') {changeO(grid, i-1, j);}if(j > 0 && grid[i][j-1] == '1'){changeO(grid, i, j-1);}if (i < grid.length - 1 && grid[i+1][j] == '1') {changeO(grid, i+1, j);}if (j < grid[i].length - 1 && grid[i][j+1] == '1') {changeO(grid, i, j+1);}return grid;}public static char[][] changeZ(char[][] grid,int i,int j) {grid[i][j] = 1;//修改为1if (i > 0 && grid[i-1][j] == '0') {changeZ(grid, i-1, j);}if(j > 0 && grid[i][j-1] == '0'){changeZ(grid, i, j-1);}if (i < grid.length - 1 && grid[i+1][j] == '0') {changeZ(grid, i+1, j);}if (j < grid[i].length - 1 && grid[i][j+1] == '0') {changeZ(grid, i, j+1);}return grid;}}
代码参考:https://segmentfault.com/a/1190000003753307
代码仍需改进的地方:(竟然可以同时计算岛屿与湖泊数量)
复杂度?
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