Week Training: 542 01 Matrix

最新推荐文章于 2025-07-06 20:57:34 发布

原创最新推荐文章于 2025-07-06 20:57:34 发布 · 179 阅读

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本文介绍了一种使用广度优先搜索(BFS)算法来计算矩阵中每个元素到最近0的距离的方法。通过将所有初始0的位置加入队列，并逐步扩展到相邻的1，更新其距离值，最终得到完整矩阵的距离值。

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Our purpose is to find the distance to the nearest 0 of each element in the matrix. It comes to me first that we should use BFS for each element, but how is still challenging. Since the returned value is a matrix, it's important how to store the answer correctly in position. We first push the position of every 0 in the matrix into a queue, then in the traversal step, we visit every elements up, down, left and right, judging if it is 0 or 1. If we find 1, the position of it is also pushed in, like the operation in BFS. During each visit, the distance is increased by 1 in this way.

class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
        int row = matrix.size();
        int col = matrix[0].size();
        queue<pair<int,int> > q;
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(matrix[i][j]==0){
                    q.push(make_pair(i,j));
                }
                else{
                    matrix[i][j]=INT_MAX;
                }
            }
        }
        int dist[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};
        while(!q.empty()){
            pair<int,int> p = q.front();
            q.pop();
            for(int i=0;i<4;i++){
                int r = p.first + dist[i][0];
                int c = p.second + dist[i][1];
                if(r<0||r>=row||c<0||c>=col||matrix[r][c]<=matrix[p.first][p.second]+1)
                    continue;
                q.push(make_pair(r,c));
                matrix[r][c] = matrix[p.first][p.second]+1;
            }
        }
        return matrix;
    }
};