[LeetCode刷题记录]Number of Islands

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. 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:

11110
11010
11000
00000

Answer: 1

Example 2:

11000
11000
00100
00011

Answer: 3

思路：很明显的典型采用深度优先搜索实现类似于Union—Find算法的变类。（图连通性）

为了不改变原来的grid内的值，我原本的思路是利用一个Bool值二维数组表示是否搜寻过。但是超出了Limited Time（可能是二维Bool数组构建和赋值时间太长）。采用把搜寻过的值改变为'X',改变grid。通过时间测试。

后来仔细想想用Bool二维数组的确没有必要。‘1’全变为‘X'最后依旧是可以复原的。

采用二维Bool值，没有通过Limited Time.

LeetCode Source

class Solution {
public:
    int numIslands(vector<vector<char>> &grid) {
        if(grid.size()==0)
            return 0;
        vector<vector<bool> >mark(grid.size(),vector<bool>(grid[0].size(),false));
        int count = 0;
        for(int i=0;i<grid.size();++i){
            for(int j=0;j<grid[0].size();++j){
                if(grid[i][j]=='1'&&!mark[i][j])
                dfs(grid,mark,i,j);
                ++count;
            }
        }
        return count;
    }
    
    void dfs(vector<vector<char>> grid,vector<vector<bool>>&mark,int i,int j){
        mark[i][j]=true;
            if(j!=grid[0].size()-1){
                if(grid[i][j]=='1'&&!mark[i][j+1])
                dfs(grid,mark,i,j+1);
            }
            if(j!=0){
                if(grid[i][j]=='1'&&!mark[i][j-1])
                dfs(grid,mark,i,j-1);
            }
            if(i!=grid.size()-1){
                if(grid[i][j]=='1'&&!mark[i+1][j])
                dfs(grid,mark,i+1,j);
            }
            if(i!=0){
                if(grid[i][j]=='1'&&!mark[i-1][j])
                dfs(grid,mark,i-1,j);
            }
        }
};

不采用二维bool数组标记，Ac了。

class Solution {
public:
    int numIslands(vector<vector<char>> &grid) {
        if(grid.size()==0)
            return 0;
        int count = 0;
        for(int i=0;i<grid.size();++i)
            for(int j=0;j<grid[0].size();++j){
                if(grid[i][j]=='1'){
                dfs(grid,i,j);
                ++count;
                }
            }
        return count;
    }
    
    void dfs(vector<vector<char>> &grid,int i,int j){
        grid[i][j]='X';
            if(j!=grid[0].size()-1){
                if(grid[i][j+1]=='1')
                dfs(grid,i,j+1);
            }
            if(j!=0){
                if(grid[i][j-1]=='1')
                dfs(grid,i,j-1);
            }
            if(i!=grid.size()-1){
                if(grid[i+1][j]=='1')
                dfs(grid,i+1,j);
            }
            if(i!=0){
                if(grid[i-1][j]=='1')
                dfs(grid,i-1,j);
            }
        }
};

后来看到别人代码，发现dfs还可以写的更加简洁一点。下面代码更加简洁：

    void dfs(vector<vector<char>> &grid,int i,int j){
        if(i<0||j<0||i>grid.size()-1||j>grid[0].size()-1||grid[i][j]!='1')
            return;
        grid[i][j]='X';
        dfs(grid,i,j+1);
        dfs(grid,i,j-1);
        dfs(grid,i+1,j);
        dfs(grid,i-1,j);
            
    }