算法刷题记录-图(LeetCode)
994. Rotting Oranges
思路 BFS
以腐烂的橘子作为起始点，使用BFS逐级向外拓展，并时刻使用cnt记录良好的橘子的数量。
代码
class Solution {
public:
int directions[4][2] = {{0, 1},
{0, -1},
{1, 0},
{-1, 0}};
int orangesRotting(vector<vector<int>> &grid) {
int cnt = 0;
deque<vector<int>> deque1;
int m=(int)grid.size();
int n=(int)grid[0].size();
for (int i = 0; i < grid.size(); ++i) {
for (int j = 0; j < grid[0].size(); ++j) {
if (grid[i][j] == 1) {
cnt++;
} else if (grid[i][j] == 2) {
deque1.push_back({i, j, 0});
}
}
}
if (cnt==0){
return 0;
}
while (!deque1.empty()) {
auto curr = deque1.back();
deque1.pop_back();
int time=curr[2];
for (auto d:directions) {
int x=curr[0]+d[0];
int y=curr[1]+d[1];
if (x>=0&&x<m&&y>=0&&y<n&&grid[x][y]==1){
cnt--;
deque1.push_front({x,y,time+1});
grid[x][y]=2;
if (cnt==0){
return time+1;
}
}
}
}
return -1;
}
};
1020. Number of Enclaves
思路 DFS
首先对边缘使用dfs算法将所有的1变为0，然后统计剩余的1的数量即可。本算法存在优化空间，其实只要第一次清除完边缘，剩下每次碰到1，+1即可。
代码
class Solution {
static final int[][] DIRECTIONS={{1,0},{-1,0},{0,1},{0,-1}};
int curr_cnt=0;
public int numEnclaves(int[][] grid) {
clearEdges(grid);
int ans=0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j <grid[0].length ; j++) {
if (grid[i][j]==1){
curr_cnt=0;
dfsCounting(grid,i,j);
ans+=curr_cnt;
}
}
}
return ans;
}
public void clearEdges(int[][] grid){
for (int i = 0; i < grid[0].length; i++) {
if (grid[0][i]==1){
dfsClear(grid,0,i);
}
if (grid[grid.length-1][i]==1){
dfsClear(grid,grid.length-1,i);
}
}
for (int i = 1; i < grid.length-1; i++) {
if (grid[i][0]==1){
dfsClear(grid,i,0);
}
if (grid[i][grid[0].length-1]==1){
dfsClear(grid,i,grid[0].length-1);
}
}
}
public void dfsClear(int[][] grid,int r,int c){
if (r>=grid.length||c>=grid[0].length||r<0||c<0||grid[r][c]==0){
return ;
}
grid[r][c]=0;
for (var d:DIRECTIONS){
dfsClear(grid,r+d[0],c+d[1]);
}
}
public void dfsCounting(int[][] grid,int r,int c){
if (r>=grid.length||c>=grid[0].length||r<0||c<0||grid[r][c]==0){
return ;
}
grid[r][c]=0;
curr_cnt++;
for (var d:DIRECTIONS){
dfsCounting(grid,r+d[0],c+d[1]);
}
}
}
1222. Queens That Can Attack the King
思路 Hash+DFS
对8个方向依次遍历，只要存在可以攻击的棋子即立即返回。
代码
class Solution {
public static class Coordinate{
public int x;
public int y;
@Override
public boolean equals(Object obj) {
if (this==obj){
return true;
}
if (obj!=null&&getClass()!=obj.getClass()){
return false;
}
Coordinate other=(Coordinate) obj;
return other.x==x&&other.y==y;
}
public Coordinate(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int hashCode() {
return Objects.hash(x,y);
}
}
List<List<Integer>> ans=new ArrayList<>();
HashSet<Coordinate> queens=new HashSet<>();
public List<List<Integer>> queensAttacktheKing(int[][] _queens, int[] king) {
for(var q:_queens){
queens.add(new Coordinate(q[0],q[1]));
}
findOneDirection(king,1,1);
findOneDirection(king,1,-1);
findOneDirection(king,-1,1);
findOneDirection(king,0,-1);
findOneDirection(king,-1,0);
findOneDirection(king,0,1);
findOneDirection(king,1,0);
findOneDirection(king,-1,-1);
return ans;
}
public void findOneDirection(int[] king ,int r,int c){
int curr_r=king[0]+r;
int curr_c=king[1]+c;
while (true){
if (curr_r<0||curr_r>=8||curr_c<0||curr_c>=8){
return;
}
if (queens.contains(new Coordinate(curr_r,curr_c))){
ans.add(List.of(curr_r,curr_c));
return;
}
curr_r+=r;
curr_c+=c;
}
}
}
2596. Check Knight Tour Configuration
思路 DFS
对于起始点，每次对八个方向进行判断查看是否有可行的下一个点。如果遍历完成则返回true。
代码
class Solution {
public:
int directions[8][2]={{2,1},{2,-1},{-2,-1},{-2,1},{1,2},{1,-2},{-1,2},{-1,-2}};
int target;
int n;
bool flag=false;
vector<vector<int>> grid;
bool checkValidGrid(vector<vector<int>>& _grid) {
grid=_grid;
n=grid.size();
if (grid[0][0]!=0){
return false;
}
target=n*n;
findGoal(1,0,0);
return flag;
}
void findGoal(int curr,int r,int c){
if (flag){
return;
}
if (curr==target){
flag= true;
return;
}
for (auto p:directions) {
int r_next=p[0]+r;
int c_next=p[1]+c;
if (r_next>=0&&r_next<n&&c_next>=0&&c_next<n&&grid[r_next][c_next]==curr){
findGoal(curr+1,r_next,c_next);
}
}
}
};