leetcode--NumberofIslands

思路：

從第一行往下對每個'1'分別對右和下相鄰爲'1'的元素做union操作，最後每個'1'都和上下左右四個相鄰的元素做了union，union and find count數即爲island數。

union and find:

有幾種實現，時間複雜度取決於find()以及union()函數，find()返回該節點屬於哪個集合，union把兩個節點對應的兩個集合進行歸併。

這裏採用最優的實現方式，find()函數搜索當前節點的根節點，union分別搜索兩個節點的根節點，並把小樹歸入大樹。

假設二維數組共有n個元素，時間複雜度爲O（nlogn)

/**
 * Created by marsares on 15/6/21.
 */
public class NumberofIslands {
    int row;
    int column;
    int count;
    int[]index;
    int[]sz;
    public int numIslands(char[][] grid) {
        if(grid==null)return 0;
        row=grid.length;
        if(row==0)return 0;
        column=grid[0].length;
        if(column==0)return 0;
        index=new int[row*column];
        sz=new int[row*column];
        for(int i=0;i<index.length;i++){
            index[i]=i;
            if(grid[i/column][i%column]=='1')count++;
        }
        for(int i=0;i<sz.length;i++){
            sz[i]=1;
        }
        for(int i=0;i<row;i++){
            for(int j=0;j<column;j++){
                if(i==row-1&&j==column-1)continue;
                else if(i==row-1){
                    if(grid[i][j]=='1'&&grid[i][j+1]=='1')union(toIndex(i, j),toIndex(i, j + 1));
                }
                else if(j==column-1){
                    if(grid[i][j]=='1'&&grid[i+1][j]=='1')union(toIndex(i,j),toIndex(i+1,j));
                }
                else{
                    if(grid[i][j]=='1'&&grid[i][j+1]=='1')union(toIndex(i,j),toIndex(i,j+1));
                    if(grid[i][j]=='1'&&grid[i+1][j]=='1')union(toIndex(i,j),toIndex(i+1,j));
                }
            }
        }
        return count;
    }
    public int find(int i){
        while(index[i]!=i){
            i=index[i];
        }
        return i;
    }
    public void union(int a,int b){
        int roota=find(a);
        int rootb=find(b);
        if(roota==rootb)return;
        if(sz[roota]>sz[rootb])index[rootb]=roota;
        else index[roota]=rootb;
        count--;
    }
    private int toIndex(int i,int j){
        return i*column+j;
    }

    public static void main(String[]args){
        NumberofIslands noi=new NumberofIslands();
        char[][]grid={{'1','1','0','0','0'},{'1','1','0','0','0'},{'0','0','1','0','0'},{'0','0','0','1','1'}};
        System.out.println(noi.numIslands(grid));
    }
}