LeetCode Hot 100：图论
200. 岛屿数量
思路 1：深度优先搜索
cpp
class Solution {
private:
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};
public:
int numIslands(vector<vector<char>>& grid) {
if (grid.empty())
return 0;
int m = grid.size(), n = m ? grid[0].size() : 0;
function<void(int, int)> dfs = [&](int x, int y) -> void {
if (x < 0 || x >= m || y < 0 || y >= n)
return;
if (grid[x][y] == '0')
return;
grid[x][y] = '0';
for (int i = 0; i < 4; i++) {
int r = x + dx[i], c = y + dy[i];
dfs(r, c);
}
};
int ans = 0;
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (grid[i][j] == '1') {
ans++;
dfs(i, j);
}
return ans;
}
};思路 2：广度优先搜索
cpp
class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
if (grid.empty())
return 0;
int m = grid.size(), n = m ? grid[0].size() : 0;
int islands = 0;
for (int r = 0; r < m; r++)
for (int c = 0; c < n; c++)
if (grid[r][c] == '1') {
islands++;
grid[r][c] = '0';
queue<pair<int, int>> neighbors;
neighbors.push({r, c});
while (!neighbors.empty()) {
auto rc = neighbors.front();
neighbors.pop();
int row = rc.first, col = rc.second;
if (row - 1 >= 0 && grid[row - 1][col] == '1') {
neighbors.push({row - 1, col});
grid[row - 1][col] = '0';
}
if (row + 1 < m && grid[row + 1][col] == '1') {
neighbors.push({row + 1, col});
grid[row + 1][col] = '0';
}
if (col - 1 >= 0 && grid[row][col - 1] == '1') {
neighbors.push({row, col - 1});
grid[row][col - 1] = '0';
}
if (col + 1 < n && grid[row][col + 1] == '1') {
neighbors.push({row, col + 1});
grid[row][col + 1] = '0';
}
}
}
return islands;
}
};994. 腐烂的橘子
思路 1：广度优先搜索
cpp
class Solution {
private:
const int dx[4] = {0, 0, 1, -1};
const int dy[4] = {1, -1, 0, 0};
public:
int orangesRotting(vector<vector<int>>& grid) {
int m = grid.size(), n = m ? grid[0].size() : 0;
queue<pair<int, int>> q;
int fresh = 0;
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1)
fresh++;
else if (grid[i][j] == 2)
q.push({i, j});
}
if (fresh == 0)
return 0;
else if (q.empty())
return -1;
int time = 0;
while (fresh > 0 && !q.empty()) {
int size = q.size();
for (int i = 0; i < size; i++) {
auto [x, y] = q.front();
q.pop();
for (int i = 0; i < 4; i++) {
int r = x + dx[i], c = y + dy[i];
if (r >= 0 && r < m && c >= 0 && c < n)
if (grid[r][c] == 1) {
fresh--;
grid[r][c] = 2;
q.push({r, c});
}
}
}
time++;
}
return fresh > 0 ? -1 : time;
}
};207. 课程表
思路 1：拓扑排序
cpp
class Solution {
public:
bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
vector<vector<int>> graph(numCourses, vector<int>()); // 邻接矩阵
vector<int> inDegree(numCourses, 0); // 入度数组
for (vector<int>& prerequisity : prerequisites) {
int from = prerequisity[1];
int to = prerequisity[0];
graph[from].push_back(to);
inDegree[to]++;
}
queue<int> q;
// 把入度为0的节点（即没有前置课程要求）放在队列中
for (int i = 0; i < inDegree.size(); i++)
if (inDegree[i] == 0)
q.push(i);
while (!q.empty()) {
int u = q.front();
q.pop();
for (int& v : graph[u]) {
inDegree[v]--;
if (inDegree[v] == 0)
q.push(v);
}
}
for (int& in : inDegree)
if (in)
return false;
return true;
}
};208. 实现 Trie (前缀树)
思路 1：字典树
cpp
class TrieNode {
public:
TrieNode* childNode[26];
bool isEnd;
TrieNode() : isEnd(false) {
for (int i = 0; i < 26; i++)
childNode[i] = nullptr;
}
};
class Trie {
private:
TrieNode* root;
public:
Trie() : root(new TrieNode()) {}
// 向字典树插入一个词
void insert(string word) {
TrieNode* node = root;
for (char& c : word) {
if (node->childNode[c - 'a'] == nullptr)
node->childNode[c - 'a'] = new TrieNode();
node = node->childNode[c - 'a'];
}
node->isEnd = true;
}
// 判断字典树里是否有一个词
bool search(string word) {
TrieNode* node = root;
for (char& c : word) {
if (node == nullptr)
break;
node = node->childNode[c - 'a'];
}
return node ? node->isEnd : false;
}
// 判断字典树是否有一个以词开始的前缀
bool startsWith(string prefix) {
TrieNode* node = root;
for (char& c : prefix) {
if (node == nullptr)
break;
node = node->childNode[c - 'a'];
}
return node != nullptr;
}
};
/**
* Your Trie object will be instantiated and called as such:
* Trie* obj = new Trie();
* obj->insert(word);
* bool param_2 = obj->search(word);
* bool param_3 = obj->startsWith(prefix);
*/