LeetCode

Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.

The distance between two adjacent cells is 1.

Example 1: 
Input:

0 0 00 1 00 0 0

Output:

0 0 00 1 00 0 0

Example 2: 
Input:

0 0 00 1 01 1 1

Output:

0 0 00 1 01 2 1

Note:

1. The number of elements of the given matrix will not exceed 10,000.
2. There are at least one 0 in the given matrix.
3. The cells are adjacent in only four directions: up, down, left and right.

Subscribe to see which companies asked this question.

题意：0/1矩阵，对于每个1求出离他最近的0需要走多少步

思路：就一个简单的bfs，纯的，裸的，特别裸。。。

心情好复杂啊。。。搞清楚了以后觉得自己是个zz，这么简单的面试题我居然写了暴力。。。（求面试官心理阴影面积T T）

class Solution {public:    struct node {        int x, y;        node() {}        node (int _x, int _y) {            x = _x;            y = _y;        }    };    vector<vector<int>> updateMatrix(vector<vector<int>>& a) {        int c = a.size(), r = a[0].size();        vector<vector<int>> ans(c, vector<int>(r, r + c));        queue<node> m;        for (int i = 0; i < c; i++) {            for (int j = 0; j < r; j++) {                if (a[i][j] == 0) {                    m.push(node(i, j));                    ans[i][j] = 0;                }            }        }        int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1};        while (!m.empty()) {            node now = m.front();            m.pop();            for (int i = 0; i < 4; i++) {                int nx = now.x + dir[i][0];                int ny = now.y + dir[i][1];                while (nx >=0 && nx < c && ny >= 0 && ny < r && ans[nx][ny] == r + c) {                    m.push(node(nx, ny));                    ans[nx][ny] = ans[now.x][now.y] + 1;                }            }        }        return ans;    }};

啊，再贴个男票写的~

vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {    int n = matrix.size(),m = matrix[0].size();    vector<vector<int>> res(n,vector<int>(m,n+m));    queue<pair<int,int>> Q;    for(int i=0;i<n;++i)        for(int j=0;j<m;++j)            if(matrix[i][j] == 0) Q.push(make_pair(i,j)),res[i][j] = 0;    int dx[] = {0,0,1,-1};    int dy[] = {1,-1,0,0};        while(!Q.empty()) {        auto u = Q.front(); Q.pop();        for(int i=0;i<4;++i){            int x = u.first + dx[i],y = u.second + dy[i];            if(x >= 0 && y >= 0 && x < n && y < m && res[x][y] == m+n){                res[x][y] = res[u.first][u.second] + 1;                Q.push(make_pair(x,y));            }        }    }    return res;}