[LeetCode] 934. Shortest Bridge 最短的桥梁

In a given 2D binary array `A`, there are two islands.  (An island is a 4-directionally connected group of `1`s not connected to any other 1s.)

Now, we may change 0s to 1s so as to connect the two islands together to form 1 island.

Return the smallest number of 0s that must be flipped.  (It is guaranteed that the answer is at least 1.)

Example 1:

Input: [[0,1],[1,0]]
Output: 1

Example 2:

Input: [[0,1,0],[0,0,0],[0,0,1]]
Output: 2

Example 3:

Input: [[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

Note:

1. 1 <= A.length = A[0].length <= 100
2. A[i][j] == 0 or A[i][j] == 1

这道题说是有一个只有0和1的二维数组，其中连在一起的1表示岛屿，现在假定给定的数组中一定有两个岛屿，问最少需要把多少个0变成1才能使得两个岛屿相连。在 LeetCode 中关于岛屿的题目还不少，但是万变不离其宗，核心都是用 DFS 或者 BFS 来解，有些还可以用联合查找 Union Find 来做。这里要求的是最小值，首先预定了一个 BFS，这就相当于洪水扩散一样，一圈一圈的，用的就是 BFS 的层序遍历。好，现在确定了这点后，再来想，这里并不是从某个点开始扩散，而是要从一个岛屿开始扩散，那么这个岛屿的所有的点都是 BFS 的起点，都是要放入到 queue 中的，所以要先来找出一个岛屿的所有点。找的方法就是遍历数组，找到第一个1的位置，然后对其调用 DFS 或者 BFS 来找出所有相连的1，先来用 DFS 的方法，对第一个为1的点调用递归函数，将所有相连的1都放入到一个队列 queue 中，并且将该点的值改为2，然后使用 BFS 进行层序遍历，每遍历一层，结果 res 都增加1，当遇到1时，直接返回 res 即可，参见代码如下：
解法一：

class Solution {
public:
    int shortestBridge(vector>& A) {
        int res = 0, n = A.size(), startX = -1, startY = -1;
        queue q;
        vector dirX{-1, 0, 1, 0}, dirY = {0, 1, 0, -1};
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (A[i][j] == 0) continue;
                startX = i; startY = j;
                break;
            }
            if (startX != -1) break;
        }
        helper(A, startX, startY, q);
        while (!q.empty()) {
            for (int i = q.size(); i > 0; --i) {
                int t = q.front(); q.pop();
                for (int k = 0; k < 4; ++k) {
                    int x = t / n + dirX[k], y = t % n + dirY[k];
                    if (x < 0 || x >= n || y < 0 || y >= n || A[x][y] == 2) continue;
                    if (A[x][y] == 1) return res;
                    A[x][y] = 2;
                    q.push(x * n + y);
                }
            }
            ++res;
        }
        return res;
    }
    void helper(vector>& A, int x, int y, queue& q) {
        int n = A.size();
        if (x < 0 || x >= n || y < 0 || y >= n || A[x][y] == 0 || A[x][y] == 2) return;
        A[x][y] = 2;
        q.push(x * n + y);
        helper(A, x + 1, y, q);
        helper(A, x, y + 1, q);
        helper(A, x - 1, y, q);
        helper(A, x, y - 1, q);
    }
};

我们也可以使用 BFS 来找出所有相邻的1，再加上后面的层序遍历的 BFS，总共需要两个 BFS，注意这里第一个 BFS 不需要是层序遍历的，而第二个 BFS 是必须层序遍历，可以对比一下看一下这两种写法有何不同，参见代码如下：
解法二：

class Solution {
public:
    int shortestBridge(vector>& A) {
        int res = 0, n = A.size();
        queue q, que;
        vector dirX{-1, 0, 1, 0}, dirY = {0, 1, 0, -1};
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (A[i][j] == 0) continue;
                A[i][j] = 2;
                que.push(i * n + j);
                break;
            }
            if (!que.empty()) break;
        }
        while (!que.empty()) {
            int t = que.front(); que.pop();
            q.push(t);
            for (int k = 0; k < 4; ++k) {
                int x = t / n + dirX[k], y = t % n + dirY[k];
                if (x < 0 || x >= n || y < 0 || y >= n || A[x][y] == 0 || A[x][y] == 2) continue;
                A[x][y] = 2;
                que.push(x * n + y);
            }
        }
        while (!q.empty()) {
            for (int i = q.size(); i > 0; --i) {
                int t = q.front(); q.pop();
                for (int k = 0; k < 4; ++k) {
                    int x = t / n + dirX[k], y = t % n + dirY[k];
                    if (x < 0 || x >= n || y < 0 || y >= n || A[x][y] == 2) continue;
                    if (A[x][y] == 1) return res;
                    A[x][y] = 2;
                    q.push(x * n + y);
                }
            }
            ++res;
        }
        return res;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/934

类似题目：

Making A Large Island

Number of Distinct Islands II

Max Area of Island

Number of Distinct Islands

Island Perimeter

Number of Islands II

Number of Islands

参考资料：

https://leetcode.com/problems/shortest-bridge/

https://leetcode.com/problems/shortest-bridge/discuss/189315/Java-DFS%2BBFS-traverse-the-2D-array-once

https://leetcode.com/problems/shortest-bridge/discuss/189293/C%2B%2B-BFS-Island-Expansion-%2B-UF-Bonus

https://leetcode.com/problems/shortest-bridge/discuss/189321/Java-DFS-find-the-island-greater-BFS-expand-the-island

[LeetCode All in One 题目讲解汇总(持续更新中...)](https://www.cnblogs.com/grandyang/p/4606334.html)