leetcode講解--695. Max Area of Island

題目

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.

題目地址

講解

一看這題我首先想到的就是並查集，只不過咱們須要稍做改動，咱們須要知道每一個並查集的元素大小，因此我加入了一個size屬性，而爲了節省空間去掉了value屬性。另外，咱們還須要一個空間來存儲DS結點，並且最好是很快的取出，因此我使用了hashmap來存，但key是一個整數元組（tuple），但java不能很方便的使用tuple，因此我直接變相的使用字符串來實現Key。

java代碼

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++;
            }
        }
    }
}