from： https://leetcode.cn/studyplan/top-100-liked/
bfs 具有 边权为1 的最短路性质
拓扑排序，入度
Trie树， 高效存储 字符串【见鬼，不知道为什么写错，需要掌握熟练度】
200. 岛屿数量【dfs / bfs】
dfs 写法，比较简洁
class Solution {
public:
int dx[4] = {-1,0,1,0}, dy[4] = {0,1,0,-1};
int n, m;
int numIslands(vector<vector<char>>& grid) {
n = grid.size(), m = grid[0].size();
int cnt = 0;
for(int i = 0;i < n;i ++ ){
for(int j = 0;j < m;j ++ ){
if(grid[i][j] == '1') {
cnt ++ ;
dfs(i, j, grid);
}
}
}
return cnt;
}
void dfs(int x, int y,vector<vector<char>>& grid){
grid[x][y] = '0';
for(int i = 0;i < 4;i ++ ){
int a = x + dx[i], b = y + dy[i];
if(a >= 0 && a < n && b >= 0 && b < m && grid[a][b] == '1')
dfs(a, b, grid);
}
};
};
bfs 写法，有最短路性质
#define x first
#define y second
class Solution {
public:
int n, m;
typedef pair<int,int> PII;
int dx[4] = {-1,0,1,0}, dy[4] = {0,1,0,-1};
int numIslands(vector<vector<char>>& grid) {
if(grid.empty() || grid[0].empty()) return 0;
n = grid.size(), m = grid[0].size();
int res = 0;
for(int i =0;i<n;i++)
for(int j=0;j<m;j++)
if(grid[i][j] == '1')
{
res ++;
bfs(i,j,grid);
}
return res;
}
void bfs(int x,int y,vector<vector<char>>& grid)
{
queue<PII> q;
q.push({x,y});
grid[x][y] = '0';
while(!q.empty())
{
auto t = q.front();
q.pop();
for(int i=0;i<4;i++)
{
int a = t.x + dx[i], b =t.y + dy[i]; // debug : 这里是新坐标的t.x 不是 x
if(a >= 0 && a < n && b >= 0 && b < m && grid[a][b] == '1')
{
grid[a][b] = '0';
q.push({a,b});
}
}
}
}
};
994. 腐烂的橘子【bfs 具有 边权为1 的最短路性质】
bfs 具有 边权为1 的最短路性质
class Solution {
public:
int orangesRotting(vector<vector<int>>& grid) {
int n = grid.size(), m = grid[0].size();
bool st[n][m];
memset(st, 0, sizeof st);
queue<pair<int,int>> q;
int dx[4] = {-1,0,1,0}, dy[4] = {0,1,0,-1};
for(int i = 0;i < n;i ++ ){
for(int j = 0; j < m;j ++ ){
if(grid[i][j] == 2) {
q.push({i, j});
st[i][j] = true;
}
}
}
int res = 0;
while(q.size()){
int k = q.size(); // debug: int k, 写成n 和 前面命名重复了！
res ++ ;
while(k -- ){
auto t = q.front();
q.pop();
for(int i = 0;i < 4;i ++ ){
int a = t.first + dx[i], b = t.second + dy[i];
if(a >= 0 && a < n && b >= 0 && b < m && grid[a][b] == 1 && !st[a][b]){
q.push({a, b});
grid[a][b] = 2;
st[a][b] = true;
}
}
}
}
for(int i = 0;i < n;i ++ ){
for(int j = 0; j < m;j ++ ){
if(grid[i][j] == 1) {
return -1;
}
}
}
if(res == 0) return 0;
return res - 1;
}
};
207. 课程表【拓扑排序】
拓扑排序
class Solution {
public:
bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
// 拓扑排序
int d[numCourses];
memset(d, 0, sizeof d);
vector<int> g[numCourses];
for(auto &c : prerequisites) {
int a = c[0], b = c[1];
g[a].push_back(b);
d[b] ++ ;
}
queue<int> q;
for(int i = 0;i < numCourses;i ++ ){
if(d[i] == 0) q.push(i);
}
while(q.size()){
int t = q.front();
q.pop();
for(auto to : g[t]){
d[to] -- ;
if(d[to] == 0) q.push(to);
}
}
for(int i = 0;i < numCourses;i ++ ){
if(d[i] != 0) return false;
}
return true;
}
};
208. 实现 Trie (前缀树)【模板题】
模板题
数组写法，简洁，需要注意开的数组空间 N * 结点文章来源：https://www.toymoban.com/news/detail-635154.html
const int N = 30010;
int tr[N * 26][26], idx;
int cnt[N * 26];
class Trie {
public:
Trie() {
idx = 0;
memset(tr, 0, sizeof tr);
memset(cnt, 0, sizeof cnt);
}
void insert(string word) {
int p = 0;
for(auto c : word){
int u = c - 'a';
if(!tr[p][u]) tr[p][u] = ++ idx;
p = tr[p][u];
}
cnt[p] ++ ;
}
bool search(string word) {
int p = 0;
for(auto c : word){
int u = c - 'a';
if(!tr[p][u]) return false;
p = tr[p][u];
}
return cnt[p] > 0;
}
bool startsWith(string prefix) {
int p = 0;
for(auto c : prefix){
int u = c - 'a';
if(!tr[p][u]) return false;
p = tr[p][u];
}
return true;
}
};
/**
* 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);
*/
指针写法文章来源地址https://www.toymoban.com/news/detail-635154.html
class Trie {
public:
struct Node
{
bool is_end;
Node *son[26];
Node()
{
is_end = false;
for(int i=0;i<26;i++) son[i] = NULL;
}
}*root;
/** Initialize your data structure here. */
Trie() {
root = new Node();
}
/** Inserts a word into the trie. */
void insert(string word) {
auto *p = root;
for(auto c : word)
{
int u = c - 'a';
if(p->son[u] == NULL) p->son[u] = new Node();
p = p->son[u];
}
p->is_end = true;
}
/** Returns if the word is in the trie. */
bool search(string word) {
auto *p = root;
for(auto c : word)
{
int u = c - 'a';
if(p->son[u] == NULL) return false;
p = p->son[u];
}
return p->is_end;
}
/** Returns if there is any word in the trie that starts with the given prefix. */
bool startsWith(string prefix) {
auto *p = root;
for(auto c : prefix)
{
int u = c - 'a';
if(p->son[u] == NULL) return false;
p = p->son[u];
}
return true;
}
};
/**
* 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);
*/
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