做易买网网站项目心得体会百度sem代运营

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01 矩阵

• 01 矩阵

class Solution {
public:vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};int m = mat.size(), n = mat[0].size();vector<vector<int>> ret(m, vector<int>(n, -1));queue<pair<int, int>> q;for (int i = 0; i < m; i++)for (int j = 0; j < n; j++)if (mat[i][j] == 0){q.push({i, j});ret[i][j] = 0;}while (q.size()){auto [a, b] = q.front();q.pop();for (int i = 0; i < 4; i++){int x = a + dx[i], y = b + dy[i];if (x >= 0 && x < m && y >= 0 && y < n && ret[x][y] == -1){ret[x][y] = ret[a][b] + 1;q.push({x, y});}}}return ret;}
};

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飞地的数量

• 飞地的数量

class Solution {
public:int numEnclaves(vector<vector<int>>& grid) {int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};int m = grid.size(), n = grid[0].size();queue<pair<int, int>> q;bool used[501][501] = {};for (int i = 0; i < m; i++)for (int j = 0; j < n; j++)if (i == 0 || i == m - 1 || j == 0 || j == n - 1)if (grid[i][j] == 1) {q.push({i, j});used[i][j] = true;}while (q.size()){auto [a, b] = q.front();q.pop();for (int i = 0; i < 4; i++){int x = a + dx[i], y = b + dy[i];if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] && !used[x][y]){used[x][y] = true;q.push({x, y});}}}int ret = 0;for (int i = 0; i < m; i++)for (int j = 0; j < n; j++)if (grid[i][j] && !used[i][j])ret++;return ret;}
};

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地图中的最高点

• 地图中的最高点

class Solution {
public:vector<vector<int>> highestPeak(vector<vector<int>>& isWater) {int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};int m = isWater.size(), n = isWater[0].size();queue<pair<int, int>> q;vector<vector<int>> ret(m, vector<int>(n, -1));for (int i = 0; i < m; i++)for (int j = 0; j < n; j++)if (isWater[i][j] == 1){q.push({i, j});ret[i][j] = 0;}while (q.size()){auto [a, b] = q.front();q.pop();for (int i = 0; i < 4; i++){int x = a + dx[i], y = b + dy[i];if (x >= 0 && x < m && y >= 0 && y < n && ret[x][y] == -1){ret[x][y] = ret[a][b] + 1;q.push({x, y});}}}return ret;}
};

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地图分析

• 地图分析

class Solution {
public:int maxDistance(vector<vector<int>>& grid) {int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};int n = grid.size();queue<pair<int, int>> q;bool used[101][101] = {};for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)if (grid[i][j] == 1)q.push({i, j});if (q.size() == n * n || q.size() == 0) return -1;int ret = -1;while (q.size()){ret++;int sz = q.size();while (sz--){auto [a, b] = q.front();q.pop();for (int i = 0; i < 4; i++){int x = a + dx[i], y = b + dy[i];if (x >= 0 && x < n && y >= 0 && y < n && !grid[x][y] && !used[x][y]){used[x][y] = true;q.push({x, y});}}}}return ret;}
};

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腐烂的苹果

• 腐烂的苹果

class Solution {
public:int rotApple(vector<vector<int> >& grid) {int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};int m = grid.size(), n = grid[0].size();queue<pair<int, int>> q;for (int i = 0; i < m; i++)for (int j = 0; j < n; j++)if (grid[i][j] == 2)q.push({i, j});int min = -1;while (q.size()){min++;int sz = q.size();while (sz--){auto [a, b] = q.front();q.pop();for (int k = 0; k < 4; k++){int x = a + dx[k], y = b + dy[k];if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1){grid[x][y] = 2;q.push({x, y});}}}}for (int i = 0; i < m; i++)for (int j = 0; j < n; j++)if (grid[i][j] == 1) return -1;return min;}
};

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