934 Shortest Bridge 最短的橋
Description:
You are given an n x n binary matrix grid where 1 represents land and 0 represents water.
An island is a 4-directionally connected group of 1's not connected to any other 1's. There are exactly two islands in grid.
You may change 0's to 1's to connect the two islands to form one island.
Return the smallest number of 0's you must flip to connect the two islands.
Example:
Example 1:
Input: grid = [[0,1],[1,0]]
Output: 1
Example 2:
Input: grid = [[0,1,0],[0,0,0],[0,0,1]]
Output: 2
Example 3:
Input: grid = [[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]]
Output: 1
Constraints:
n == grid.length == grid[i].length
2 <= n <= 100
grid[i][j] is either 0 or 1.
There are exactly two islands in grid.
題目描述:
在給定的二維二進制數組 A 中，存在兩座島。（島是由四面相連的 1 形成的一個最大組。）
現在，我們可以將 0 變爲 1，以使兩座島連接起來，變成一座島。
返回必須翻轉的 0 的最小數目。（可以保證答案至少是 1 。）
示例 :
示例 1：
輸入：A = [[0,1],[1,0]]
輸出：1
示例 2：
輸入：A = [[0,1,0],[0,0,0],[0,0,1]]
輸出：2
示例 3：
輸入：A = [[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]]
輸出：1
提示:
2 <= A.length == A[0].length <= 100
A[i][j] == 0 或 A[i][j] == 1
思路:
DFS + BFS
先用 DFS 的方式找到第一座島, 並將第一座島加入隊列
然後用 BFS 的方式搜索第一座島到第二座島的最短距離
可以將原島嶼值改成 2 標記訪問過的位置
時間複雜度爲 O(n ^ 2), 空間複雜度爲 O(n ^ 2)
代碼:
C++:
class Solution
{
public:
int shortestBridge(vector<vector<int>>& grid)
{
queue<int> q;
int flag = 0, n = grid.size(), result = -1;
const vector<vector<int>> d{ { 0, 1 }, { 0, -1 }, { 1, 0 }, { -1, 0 } };
for (int i = 0; i < n; i++)
{
if (flag) break;
for (int j = 0; j < n; j++)
{
if (grid[i][j])
{
dfs(grid, n, i, j, q, d);
flag = true;
break;
}
}
}
while (!q.empty())
{
int s = q.size();
++result;
for (int i = 0; i < s; i++)
{
int cur = q.front();
q.pop();
for (const auto& next : d)
{
int x = next[0] + cur / n, y = next[1] + cur % n;
if (x < 0 or x >= n or y < 0 or y >= n or grid[x][y] == 2) continue;
if (grid[x][y] == 1) return result;
q.push(x * n + y);
grid[x][y] = 2;
}
}
}
return result;
}
private:
void dfs(vector<vector<int>>& grid, int n, int i, int j, queue<int>& q, const vector<vector<int>>& d)
{
if (i < 0 or i >= n or j < 0 or j >= n or grid[i][j] != 1) return;
grid[i][j] = 2;
q.push(i * n + j);
for (const auto& next : d) dfs(grid, n, i + next[0], j + next[1], q, d);
}
};
Java:
class Solution {
public int shortestBridge(int[][] grid) {
Queue<Integer> queue = new LinkedList<>();
boolean flag = false;
int n = grid.length, result = -1, d[][] = new int[][]{ { 0, 1 }, { 0, -1 }, { 1, 0 }, { -1, 0 } };
for (int i = 0; i < n; i++) {
if (flag) break;
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1) {
dfs(grid, n, i, j, queue, d);
flag = true;
break;
}
}
}
while (!queue.isEmpty()) {
int s = queue.size();
++result;
for (int i = 0; i < s; i++) {
int cur = queue.poll();
for (int[] next : d) {
int x = next[0] + cur / n, y = next[1] + cur % n;
if (x < 0 || x >= n || y < 0 || y >= n || grid[x][y] == 2) continue;
if (grid[x][y] == 1) return result;
queue.offer(x * n + y);
grid[x][y] = 2;
}
}
}
return result;
}
private void dfs(int[][] grid, int n, int i, int j, Queue<Integer> queue, int[][] d) {
if (i < 0 || i >= n || j < 0 || j >= n || grid[i][j] != 1) return;
grid[i][j] = 2;
queue.offer(i * n + j);
for (int[] next : d) dfs(grid, n, i + next[0], j + next[1], queue, d);
}
}
Python:
class Solution:
def shortestBridge(self, grid: List[List[int]]) -> int:
n, queue, flag, d, result = len(grid), deque(), False, [(-1, 0), (1, 0), (0, -1), (0, 1)], -1
def dfs(i: int, j: int) -> None:
if not -1 < i < n or not -1 < j < n or not grid[i][j] or grid[i][j] == 2:
return
grid[i][j] = 2
queue.append((i, j))
for dx, dy in d:
dfs(i + dx, j + dy)
for i in range(n):
if flag:
break
for j in range(n):
if grid[i][j]:
dfs(i, j)
flag = True
break
while queue:
s = len(queue)
result += 1
for i in range(s):
cur = queue.popleft()
for dx, dy in d:
x, y = cur[0] + dx, cur[1] + dy
if not -1 < x < n or not -1 < y < n or grid[x][y] == 2:
continue
if grid[x][y] == 1:
return result
queue.append((x, y))
grid[x][y] = 2
return result