文章目录
  1. 1. Number of Islands
    1. 1.1. 题目
    2. 1.2. 思路
    3. 1.3. 解题

Number of Islands

题目

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

DFS

解题

c++版

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class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
if(grid.empty())
return 0;
int m=grid.size();
int n=grid[0].size();
int cnt=0;
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(grid[i][j]=='1'){
dfs(grid,i,j);
cnt++;
}
}
}
return cnt;
}

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

python版

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class Solution:
# @param {character[][]} grid
# @return {integer}
def numIslands(self, grid):
if len(grid)==0 or len(grid[0])==0:
return 0
m=len(grid)
n=len(grid[0])
cnt=0
for i in range(m):
for j in range(n):
if grid[i][j]=='1':
self.dfs(grid,i,j)
cnt+=1
return cnt


def dfs(self,grid,i,j):
if i<0 or i>=len(grid) or j<0 or j>=len(grid[0]):
return
if grid[i][j]!='1':
return
grid[i][j]='X'
self.dfs(grid,i+1,j)
self.dfs(grid,i-1,j)
self.dfs(grid,i,j+1)
self.dfs(grid,i,j-1)

文章目录
  1. 1. Number of Islands
    1. 1.1. 题目
    2. 1.2. 思路
    3. 1.3. 解题