[leetcode]934. Shortest Bridge

Analysis

2019要开心鸭—— [每天刷题并不难0.0]

In a given 2D binary array A, there are two islands. (An island is a 4-directionally connected group of 1s 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.)

Explanation:

先用DFS标记island，再用BFS扩充island，从而找到bridge的数量

Implement

class Solution {public:int shortestBridge(vector<vector<int>>& A) {pair<int, int> start;m = A.size();n = A[0].size();vector<vector<int>> visit(m, vector<int>(n));queue<pair<int, int>> mq;int flag = 0;for(int i=0; i<m; i++){if(flag == 1)break;for(int j=0; j<n; j++){if(A[i][j] == 1){DFS(i, j, A, visit, mq);flag = 1;break;}}}//DFS(start.first, start.second, A, visit, mq);int res = 0;while(!mq.empty()){int sz = mq.size();while(sz-- > 0){pair<int, int> cur = mq.front();mq.pop();for(int i=0; i<4; i++){int x = cur.first+dir[i][0];int y = cur.second+dir[i][1];if(x>=0 && y>=0 && x<m && y<n && visit[x][y]==0){if(A[x][y] == 1)return res;visit[x][y] = 1;mq.emplace(x, y);}}}res++;}return -1;}void DFS(int x, int y, vector<vector<int>>& A, vector<vector<int>>& visit, queue<pair<int, int>>& mq){if(x<0 || x>=m || y<0 || y>=n || visit[x][y]==1 || A[x][y]==0)return ;visit[x][y] = 1;mq.emplace(x, y);for(int i=0; i<4; i++)DFS(x+dir[i][0], y+dir[i][1], A, visit, mq);}
private:int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};int m, n;
};

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[leetcode]934. Shortest Bridge

[leetcode]934. Shortest Bridge

Analysis

Explanation:

Implement

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