笔试面试热点dfs｜刷题打卡

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一、题目描述：

Given an m x n 2d grid map of '1's (land) and '0's (water), return the number of islands which the sum of 1‘s on the island equal S (S>0).
An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically.
You may assume all four edges of the grid are all surrounded by water.
Example 1:
Input: grid = [
["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
], S = 9
Output: 1
public int sumIslands(char[][] grid, int S){
    // implementation
}
//----------
// unit test
//----------
Tip: using Junit4.
public void sumIslandsTest(){
    char[][] grid = [
    ["1","1","1","1","0"],
    ["1","1","0","1","0"],
    ["1","1","0","0","0"],
    ["0","0","0","0","0"]
    ];
    int S = 8;
    // test implementation
}

二、思路分析：

岛屿问题的小变种，从算有几个岛屿变成算岛屿多大。

一次完整的dfs

1. 判断退出条件（非法或合法）
2. 记录已访问
3. return 递归其他点

这里非法指越界，合法指已经遍历过该点

三、AC 代码：

package com.company;

public class * {
    public int sizer;
    public int sizel;
    public int sumIslands(char[][] grid, int S){
        // implementation
        this.sizer = grid.length;
        this.sizel = grid[0].length;
        int ret = 0;
        for (int i = 0; i < grid.length; i++){
            for (int j = 0; j < grid[0].length; j++){
                int tmp = dfs(grid, i, j);
                if(tmp == S)ret++;
            }
        }
        return ret;
    }
    public int dfs(char[][] grid, int r, int l){
        //非法
        if(r >= sizer || r < 0 || l < 0 || l >= sizel){
            return 0;
        }
        //不是岛屿或已经遍历
        if(grid[r][l] != 1){
            return 0;
        }
        grid[r][l] = 2;
        return 1 + dfs(grid,r + 1, l)+ dfs(grid,r , l + 1)+ dfs(grid,r - 1, l)+ dfs(grid,r, l - 1);
    }
//----------
// unit test
//----------
//    Tip: using Junit4.
    public void sumIslandsTest(){
        char[][] grid = {
                {'1', '1', '1', '1', '0'},
                {'1', '1', '0', '1', '0'},
                {'1', '1', '0', '0', '0'},
                {'0', '0', '0', '0', '0'}
        };
        int S = 8;
        // test implementation
        // 由于S = 8，而岛屿等于9 因此返回0
        System.out.println(sumIslands(grid,S));
    }

    public static void main(String[] args) {
        * v = new *();
        v.sumIslandsTest();
    }
}

四、总结：

由于一直对回溯，递归的抵触，导致我每次遇到这类题都想着位运算，并查集的骚套路，为此吃了不少亏，希望各位不要走我老路。