Algorithms—200.Number of Islands

思路，遍历二维char数组，判断其是否在map中，如果不在map中，且值为‘1’；那么将计数+1，并将点加入map中，然后递归查询周围4个点（考虑边界）情况将

public class Solution {    public int numIslands(char[][] grid) {int il = 0;Map<String, String> map = new HashMap<String, String>();for (int i = 0; i < grid.length; i++) {for (int j = 0; j < grid[i].length; j++) {if (map.get("(" + i + "," + j + ")") == null) {if (grid[i][j] == '1') {il++;map = new Solution().ss(map, i, j, grid);}}}}return il;    }   public Map<String,String> ss(Map<String,String> map,int i,int j,char[][] grid){if (map.get("(" + i + "," + j + ")") == null && grid[i][j] == '1') {int LI = grid.length;int LJ = grid[i].length;map.put("(" + i + "," + j + ")", "1");if (i > 0) {// 上边检测map = ss(map, i - 1, j, grid);}if (i < LI - 1) {// 下边检测map = ss(map, i + 1, j, grid);}if (j > 0) {// 左边检测map = ss(map, i, j - 1, grid);}if (j < LJ - 1) {// 右边检测map = ss(map, i, j + 1, grid);}}return map;}}

1的点全部加入。