Number of Islands
题目:
Given a boolean 2D matrix, find the number of islands.
example:
[1, 1, 0, 0, 0],
[0, 1, 0, 0, 1],
[0, 0, 0, 1, 1],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 1]
分析:
这是一道典型的搜索,计数问题,可以用dfs和bfs两种方法来解。
解法bfs:
class Solution {
public:
/**
* @param grid a boolean 2D matrix
* @return an integer
*/
void bfs(vector<vector<bool> > &grid, int i, int j) {
queue<int> Q;
int m = grid.size();
int n = grid[0].size();
grid[i][j] = false;
int code = i * n + j;
Q.push(code);
while (!Q.empty() ) {
code = Q.front();
Q.pop();
int x = code / n;
int y = code % n;
if (x - 1 >= 0 && grid[x - 1][y] == true) {
Q.push( (x - 1) * n + y);
grid[x - 1][y] = false;
}
if (x + 1 < m && grid[x + 1][y] == true) {
Q.push( (x + 1) * n + y);
grid[x + 1][y] = false;
}
if (y - 1 >= 0 && grid[x][y - 1] == true) {
Q.push(x * n + y - 1);
grid[x][y - 1] = false;
}
if (y + 1 < n && grid[x][y + 1] == true) {
Q.push(x * n + y + 1);
grid[x][y + 1] = false;
}
}
}
int numIslands(vector<vector<bool>>& grid) {
// Write your code here
if (grid.size() == 0 || grid[0].size() == 0) {
return 0;
}
int m = grid.size();
int n = grid[0].size();
int count = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == true) {
count++;
bfs(grid, i, j);
}
}
}
return count;
}
};
解法dfs:
public:
/**
* @param grid a boolean 2D matrix
* @return an integer
*/
void dfs(vector<vector<bool> > &grid, int i, int j) {
if (i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size() ) {
return;
}
if (grid[i][j] == false) {
return;
}
grid[i][j] = false;
dfs(grid, i - 1, j);
dfs(grid, i + 1, j);
dfs(grid, i, j - 1);
dfs(grid, i, j + 1);
}
int numIslands(vector<vector<bool>>& grid) {
// Write your code here
if (grid.size() == 0 || grid[0].size() == 0) {
return 0;
}
int m = grid.size();
int n = grid[0].size();
int count = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == true) {
count++;
dfs(grid, i, j);
}
}
}
return count;
}
};