给定一个由 0 和 1 组成的矩阵,找出每一个元素到最近的 0 的距离。web
两个相邻元素间的距离为 1 。svg
示例 1:
输入:spa
0 0 0 0 1 0 0 0 0
输出:.net
0 0 0 0 1 0 0 0 0
示例 2:
输入:code
0 0 0 0 1 0 1 1 1
输出:xml
0 0 0 0 1 0 1 2 1
注意:blog
BFStoken
把全部0
加入队列,找1
判断距离队列
class Solution { public: vector<vector<int>> updateMatrix(vector<vector<int>>& ma) { using paii = pair<int, int>; const int dir[4][2] = {0 , 1, 1, 0, 0, -1, -1, 0}; int n = ma.size(), m = ma[0].size(); queue<paii> q; for (int i = 0; i < n; ++i) for (int j = 0; j < m; ++j) { if (ma[i][j]) ma[i][j] = 0x3f3f3f3f; else q.push(paii(i, j)); } while (!q.empty()) { int x = q.front().first, y = q.front().second; q.pop(); for (int i = 0; i < 4; ++i) { int dx = x + dir[i][0], dy = y + dir[i][1]; if (dx >= 0 && dx < n && dy >= 0 && dy < m && ma[dx][dy] > ma[x][y] + 1) { q.push(paii(dx, dy)); ma[dx][dy] = ma[x][y] + 1; } } } return ma; } };
DPleetcode
和 LeetCode 64.最小路径和 基本是一致的 题解 ,这个题有四个方向而已
class Solution { public: vector<vector<int>> updateMatrix(vector<vector<int>>& ma) { int n = ma.size(), m = ma[0].size(); for (int i = 0; i < n; ++i) for (int j = 0; j < m; ++j)//左上,正上 { if (ma[i][j]) ma[i][j] = 0x3f3f3f3f; if (i) ma[i][j] = min(ma[i][j], ma[i - 1][j] + 1); if (j) ma[i][j] = min(ma[i][j], ma[i][j - 1] + 1); } for (int i = n - 1; i >= 0; --i) for (int j = m - 1; j >=0; --j)//右下,正下 { if (i != n - 1) ma[i][j] = min(ma[i][j], ma[i + 1][j] + 1); if (j != m - 1) ma[i][j] = min(ma[i][j], ma[i][j + 1] + 1); } return ma; } };