• 305 Number Of Islands II


    问题描述:

    Given a n,m which means the row and column of the 2D matrix and an array of pair A( size k). Originally, the 2D matrix is all 0 which means there is only sea in the matrix. The list pair has k operator and each operator has two integer A[i].x, A[i].y means that you can change the grid matrix[A[i].x][A[i].y] from sea to island. Return how many island are there in the matrix after each operator.

    解题思路:

    可以使用union-find(并查集)来解决这道问题。

    在本题中提供的为二维坐标,但并查集中多用一维数组,我们可以通过使用 idx = x * m + y 来唯一确定二维数组中的一个位置的一维表示方法。其中m为列数。

    再插入每一个位置的时候,检查上下左右是否为岛屿,如果是,则需要合并岛屿。

    代码:

    **
     * Definition for a point.
     * class Point {
     *     int x;
     *     int y;
     *     Point() { x = 0; y = 0; }
     *     Point(int a, int b) { x = a; y = b; }
     * }
     */
    
    public class Solution {
        /**
         * @param n: An integer
         * @param m: An integer
         * @param operators: an array of point
         * @return: an integer array
         */
        public List<Integer> numIslands2(int n, int m, Point[] operators) {
            // write your code here
            ArrayList<Integer> ret = new ArrayList<Integer>();
            
            if(operators == null || operators.length == 0)
                return ret;
            
            
            int count = 0;
            int[] parent = new int[n*m];
            int[][] direction = {{-1,0}, {1, 0},{0,-1},{0,1}};
            for(int i = 0; i < n*m; i++){
                parent[i] = -1;
            }
            for(int i = 0; i < operators.length; i++){
                
                Point curPoint = operators[i];
                int curID = calID(curPoint.x, curPoint.y, m);
                if(parent[curID] != -1){
                    ret.add(count);
                    continue;
                }
                count++;
                parent[curID] = curID;
                //check surrounds
                if(curPoint.x - 1 >= 0){
                    int adjID = calID(curPoint.x-1, curPoint.y, m);
                    if(parent[adjID] != -1){
                        count = union(parent, curID, adjID, count);
                    }
                }
                if(curPoint.y - 1 >= 0){
                    int adjID = calID(curPoint.x, curPoint.y-1, m);
                    if(parent[adjID] != -1){
                        count = union(parent, curID, adjID, count);
                    }
                }
                if(curPoint.x + 1 < n){
                    int adjID = calID(curPoint.x + 1, curPoint.y, m);
                    if(parent[adjID] != -1){
                        count = union(parent, curID, adjID, count);
                    }
                }
                if(curPoint.y + 1 < m){
                    int adjID = calID(curPoint.x, curPoint.y + 1, m);
                    if(parent[adjID] != -1){
                        count = union(parent, curID, adjID, count);
                    }
                }
                ret.add(count);
            }
            return ret;
        }
        
        private int calID(int i, int j, int row){
            return i*row + j;
        }
        
        private int find(int[] parent, int a){
            if(parent[a] == a){
                return a;
            } 
            return parent[a] = find(parent, parent[a]);
        }
        
        private int union(int[] parent, int a, int b, int count){
            int rootA = find(parent, a);
            int rootB = find(parent, b);
            if(rootA != rootB) {
                parent[rootB] = rootA;
                count--;
            }
            return count;
        }
    }

    其中,check surround中可以通过二维数组direction来进行检查

    int[][] direction = {{-1,0}, {1, 0},{0,-1},{0,1}};

    检查代码为:

    for(int[] d : direction){
                    int adjX = curPoint.x + d[0];
                    int adjY = curPoint.y + d[1];
                    
                    int adjID = calID(adjX, adjY, m);
                    
                    if( adjX >= 0 && adjX < n 
                        &&adjY >= 0 && adjY < m && parent[adjID] != -1){
                            count = union(parent, curID, adjID, count);
                        }
                }

    粗心出现过的bug _(:з」∠)_:

    1. 在计算唯一id:calID时,将x*m+y 算成了x*n+y, 导致了id不唯一 TAT
    2. 在周围检测时,在检查0的时候,用了<0而非<=0,导致边缘无法跳入代码块

    做题用的时间不多,但是debug反而用了很久。。以后一定要多多小心,仔细答题!

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  • 原文地址:https://www.cnblogs.com/yaoyudadudu/p/8721096.html
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