本身挺直观的题,但很容易出错。用set记录使用过的tickets中的index,然后每次遇到一种arrangement以后,在backtracking的if那里比一下哪个lexical order更小。但发现不用这样。因为数组本身不大,可以将它排序以后,这样满足if的第一个list就是我们要找的了。所以,用return true就可以找到了,不用继续往下找了。
卡哥用map记录飞行次数的方法更简单。他不排序,先过了一遍tickets记录每个dest的数量。这样其实更好。
PS: java的list.remove(i)的i是指index。如果要remove这个object,且list是Integertype的话,需要list.remove(new Integer(i))之类的。
class Solution {
private boolean first = true;
private List<String> tmp = new ArrayList<>();
private List<String> res = new ArrayList<>();
private Set<Integer> indexSet = new HashSet<>();
public List<String> findItinerary(List<List<String>> tickets) {
Collections.sort(tickets, new Comparator<List<String>>(){
@Override
public int compare(List<String> list1, List<String> list2){
return list1.get(1).compareTo(list2.get(1));
}
});
res.add("JFK");
// System.out.println(tickets.toString().toString());
backtracking(tickets, "JFK");
return tmp;
}
private void removeFromBack(String dest){
for(int i = res.size()-1; i >= 0; --i){
if(res.get(i).equals(dest)){
res.remove(i);
break;
}
}
}
private boolean backtracking(List<List<String>> tickets, String dest){
if(indexSet.size() == tickets.size()){
// if(first){
// first = false;
// tmp = new ArrayList<>(res);
// }else{
// boolean ans = tmp.toString().compareTo(res.toString()) <= 0;
// if(ans){
// return;
// }else{
// tmp = new ArrayList<>(res);
// }
// }
tmp = new ArrayList<>(res);
return true;
}
for(int i = 0; i < tickets.size(); ++i){
if(dest.equals(tickets.get(i).get(0)) && !indexSet.contains(i)){
indexSet.add(i);
res.add(tickets.get(i).get(1));
// System.out.println(indexSet.size() + " " + tickets.size());
// System.out.println(res.toString());
// System.out.println();
if(backtracking(tickets, tickets.get(i).get(1))) return true;
removeFromBack(tickets.get(i).get(1));
indexSet.remove(i);
}
}
return false;
}
}
class Solution {
private List<List<String>> res = new ArrayList<>();
private List<List<String>> board = new ArrayList<>();
public List<List<String>> solveNQueens(int n) {
for(int i = 0; i < n; ++i){
List<String> tmp = new ArrayList<>();
for(int j = 0; j < n; ++j){
tmp.add(".");
}
board.add(tmp);
}
backtracking(n, 0);
return res;
}
private void backtracking(int n, int row){
if(row == n){
List<String> newBoard = toBoard(n);
res.add(newBoard);
return;
}
for(int i = 0; i < n; i++){
if(isValid(row, i, n)){
board.get(row).set(i, "Q");
row++;
backtracking(n, row);
row--;
board.get(row).set(i, ".");
}
}
}
private List<String> toBoard(int n){
List<String> str = new ArrayList<>();
for(int i = 0; i < n; ++i){
String toAdd = "";
for(int j = 0; j < n; ++j){
if(board.get(i).get(j).equals("Q")){
toAdd += "Q";
}else{
toAdd += ".";
}
}
str.add(toAdd);
}
return str;
}
private boolean isValid(int row, int col, int n){
for (int i = 0; i < row; i++) {
if (board.get(i).get(col).equals("Q")) {
return false;
}
}
for (int i = 0; i < row; i++) {
if (board.get(row).get(i).equals("Q")) {
return false;
}
}
for (int i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--) {
if (board.get(i).get(j).equals("Q")) {
return false;
}
}
for(int i = row + 1, j = col + 1; i < n && j < n; i++, j++) {
if (board.get(i).get(j).equals("Q")) {
return false;
}
}
for(int i = row + 1, j = col - 1; i < n && j >= 0; i++, j--) {
if (board.get(i).get(j).equals("Q")) {
return false;
}
}
for(int i = row - 1, j = col + 1; i >= 0 && j < n; i--, j++) {
if (board.get(i).get(j).equals("Q")) {
return false;
}
}
// System.out.println(Arrays.toString(board.toArray()));
// System.out.println("row: " + row + " col: " + col);
return true;
}
}
class Solution {
private boolean isValid(int row, int col, char num, char[][] board){
for (int i = 0; i < 9; ++i) {
if (board[i][col] == num)
return false;
}
for (int j = 0; j < 9; ++j) {
if (board[row][j] == num)
return false;
}
int startRow = (row / 3) * 3;
int startCol = (col / 3) * 3;
for (int i = startRow; i < startRow + 3; ++i) {
for (int j = startCol; j < startCol + 3; ++j) {
if (board[i][j] == num)
return false;
}
}
return true;
}
private boolean backtracking(char[][] board){
for(int i = 0; i < 9; ++i){
for(int j = 0; j < 9; ++j){
if(board[i][j] == '.'){
for(char k = '1'; k <= '9'; ++k){
if(isValid(i, j, k, board)){
board[i][j] = k;
if(backtracking(board)) return true;
board[i][j] = '.';
}
}
return false;
}
}
}
return true;
}
public void solveSudoku(char[][] board) {
backtracking(board);
}
}
本文介绍了三个使用Java解决的算法问题:1.如何重构机票路线以得到可行的行程;2.实现N皇后问题的解决方案,展示如何放置皇后以避免冲突;3.解决数独问题的回溯算法。每个问题都涉及到了回溯法在解决约束满足问题中的应用。
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