[leetcode] 542. 01 Matrix (Medium)
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给予一个矩阵,矩阵有1有0,计算每一个1到0需要走几步,只能走上下左右。
解法一:
利用dp,从左上角遍历一遍,再从右下角遍历一遍,dp存储当前位置到0的最短距离。
十分粗心的搞错了col和row,改了半天…………
Runtime: 132 ms, faster than 98.88% of C++ online submissions for 01 Matrix.
class Solution { public: vector<vector<int>> updateMatrix(vector<vector<int>> &matrix) { if (matrix.size() == 0 || matrix[0].size() == 0) return matrix; int n; int m; n = matrix.size(); m = matrix[0].size(); int rangeNum = n + m; vector<vector<int>> dis(n, vector<int>(m, 0)); for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) { if (matrix[i][j] == 0) dis[i][j] = 0; else { int up = (i > 0) ? dis[i - 1][j] : rangeNum; int left = (j > 0) ? dis[i][j - 1] : rangeNum; dis[i][j] = min(left, up) + 1; } } for (int i = n - 1; i >= 0; i--) for (int j = m - 1; j >= 0; j--) { if (matrix[i][j] == 0) dis[i][j] = 0; else { int right = (j + 1) < m ? dis[i][j + 1] : rangeNum; int down = (i + 1) < n ? dis[i + 1][j] : rangeNum; dis[i][j] = min(min(right, down) + 1, dis[i][j]); } } return dis; } };
解法二:
BFS
class Solution { private: bool isValid(int m, int n, int x, int y) { return x >= 0 && y >= 0 && x < m && y < n; } int getShortestDistance(int m, int n, int x, int y, vector<vector<int>> &distance) { int result = distance[x][y]; if (isValid(m, n, x, y + 1) && distance[x][y + 1] != INT_MAX) { result = min(result, 1 + distance[x][y + 1]); } if (isValid(m, n, x, y - 1) && distance[x][y - 1] != INT_MAX) { result = min(result, 1 + distance[x][y - 1]); } if (isValid(m, n, x + 1, y) && distance[x + 1][y] != INT_MAX) { result = min(result, 1 + distance[x + 1][y]); } if (isValid(m, n, x - 1, y) && distance[x - 1][y] != INT_MAX) { result = min(result, 1 + distance[x - 1][y]); } return result; } public: vector<vector<int>> updateMatrix(vector<vector<int>> &matrix) { int m = matrix.size(); int n = matrix[0].size(); vector<vector<int>> distance(m, vector<int>(n, INT_MAX)); queue<pair<int, int>> visit; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (matrix[i][j] == 0) { distance[i][j] = 0; visit.push(make_pair(i, j + 1)); visit.push(make_pair(i, j - 1)); visit.push(make_pair(i + 1, j)); visit.push(make_pair(i - 1, j)); } } } while (!visit.empty()) { pair<int, int> cur = visit.front(); visit.pop(); int x = cur.first; int y = cur.second; if (isValid(m, n, x, y)) { int shortestD = getShortestDistance(m, n, x, y, distance); if (shortestD < distance[x][y]) { distance[x][y] = shortestD; visit.push(make_pair(x, y + 1)); visit.push(make_pair(x, y - 1)); visit.push(make_pair(x + 1, y)); visit.push(make_pair(x - 1, y)); } } } return distance; } };
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