leetcode讲解--695. Max Area of Island

题目

Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:

[[0,0,1,0,0,0,0,1,0,0,0,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,1,1,0,1,0,0,0,0,0,0,0,0],
 [0,1,0,0,1,1,0,0,1,0,1,0,0],
 [0,1,0,0,1,1,0,0,1,1,1,0,0],
 [0,0,0,0,0,0,0,0,0,0,1,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,0,0,0,0,0,0,1,1,0,0,0,0]]

Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

[[0,0,0,0,0,0,0,0]]

Given the above grid, return 0.
Note: The length of each dimension in the given grid does not exceed 50.

题目地址

讲解

一看这题我首先想到的就是并查集，只不过咱们须要稍做改动，咱们须要知道每一个并查集的元素大小，因此我加入了一个size属性，而为了节省空间去掉了value属性。另外，咱们还须要一个空间来存储DS结点，并且最好是很快的取出，因此我使用了hashmap来存，但key是一个整数元组（tuple），但java不能很方便的使用tuple，因此我直接变相的使用字符串来实现Key。

java代码

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
        Map<String, DS> map = new HashMap<>();
        for(int i=0;i<grid.length;i++){
            for(int j=0;j<grid[i].length;j++){
                if(grid[i][j]==1){
                    DS ds = new DS(1);
                    map.put(i+","+j, ds);
                    if(i-1>=0 && grid[i-1][j]==1){
                        ds.union(map.get((i-1)+","+j), ds);
                    }
                    if(j-1>=0 && grid[i][j-1]==1){
                        ds.union(map.get(i+","+(j-1)), ds);
                    }
                }
            }
        }
        int result = 0;
        for(String s:map.keySet()){
            int size = map.get(s).find().size;
            if(result<size){
                result = size;
            }
        }
        return result;
    }
    
    class DS{
        // private int value;
        private DS parent;
        private int rank;
        private int size;
        public DS(int value){
            // this.value = value;
            rank = 0;
            parent = this;
            size = 1;
        }
        public DS find(){
            if(this.parent!=this){
                this.parent = this.parent.find();
            }
            return this.parent;
        }
        
        public void union(DS x, DS y){
            DS xParent = x.find();
            DS yParent = y.find();
            if(xParent==yParent){
                return;
            }
            if(xParent.rank>yParent.rank){
                yParent.parent = xParent;
                xParent.size += yParent.size;
            }else if(xParent.rank<yParent.rank){
                xParent.parent = yParent;
                yParent.size += xParent.size;
            }else{
                xParent.parent = yParent;
                yParent.size += xParent.size;
                yParent.rank++;
            }
        }
    }
}