332.重新安排行程

感觉挺麻烦，需要记录映射关系来处理死循环问题

class Solution {private LinkedList<String> res;private LinkedList<String> path = new LinkedList<>();public List<String> findItinerary(List<List<String>> tickets) {Collections.sort(tickets, (a, b) -> a.get(1).compareTo(b.get(1)));path.add("JFK");boolean[] used = new boolean[tickets.size()];backTracking((ArrayList) tickets, used);return res;}public boolean backTracking(ArrayList<List<String>> tickets, boolean[] used) {if (path.size() == tickets.size() + 1) {res = new LinkedList(path);return true;}for (int i = 0; i < tickets.size(); i++) {if (!used[i] && tickets.get(i).get(0).equals(path.getLast())) {path.add(tickets.get(i).get(1));used[i] = true;if (backTracking(tickets, used)) {return true;}used[i] = false;path.removeLast();}}return false;}
}

51. N皇后

回溯好写，检查n皇后的代码之前没见过，比较像五子棋

class Solution {List<List<String>> res = new ArrayList<>();public List<List<String>> solveNQueens(int n) {char[][] chessboard = new char[n][n];for (char[] c : chessboard) {Arrays.fill(c, '.');}backTrack(n, 0, chessboard);return res;}public void backTrack(int n, int row, char[][] chessboard) {if (row == n) {res.add(Array2List(chessboard));return;}for (int col = 0;col < n; ++col) {if (isValid (row, col, n, chessboard)) {chessboard[row][col] = 'Q';backTrack(n, row+1, chessboard);chessboard[row][col] = '.';}}}public List Array2List(char[][] chessboard) {List<String> list = new ArrayList<>();for (char[] c : chessboard) {list.add(String.copyValueOf(c));}return list;}public boolean isValid(int row, int col, int n, char[][] chessboard) {// 检查列for (int i=0; i<row; ++i) { // 相当于剪枝if (chessboard[i][col] == 'Q') {return false;}}// 检查45度对角线for (int i=row-1, j=col-1; i>=0 && j>=0; i--, j--) {if (chessboard[i][j] == 'Q') {return false;}}// 检查135度对角线for (int i=row-1, j=col+1; i>=0 && j<=n-1; i--, j++) {if (chessboard[i][j] == 'Q') {return false;}}return true;}
}

37. 解数独

巨麻烦，要是面试遇到就认了

class Solution {public void solveSudoku(char[][] board) {solveSudokuHelper(board);}private boolean solveSudokuHelper(char[][] board){//「一个for循环遍历棋盘的行，一个for循环遍历棋盘的列，// 一行一列确定下来之后，递归遍历这个位置放9个数字的可能性！」for (int i = 0; i < 9; i++){ // 遍历行for (int j = 0; j < 9; j++){ // 遍历列if (board[i][j] != '.'){ // 跳过原始数字continue;}for (char k = '1'; k <= '9'; k++){ // (i, j) 这个位置放k是否合适if (isValidSudoku(i, j, k, board)){board[i][j] = k;if (solveSudokuHelper(board)){ // 如果找到合适一组立刻返回return true;}board[i][j] = '.';}}// 9个数都试完了，都不行，那么就返回falsereturn false;// 因为如果一行一列确定下来了，这里尝试了9个数都不行，说明这个棋盘找不到解决数独问题的解！// 那么会直接返回， 「这也就是为什么没有终止条件也不会永远填不满棋盘而无限递归下去！」}}// 遍历完没有返回false，说明找到了合适棋盘位置了return true;}/*** 判断棋盘是否合法有如下三个维度:*     同行是否重复*     同列是否重复*     9宫格里是否重复*/private boolean isValidSudoku(int row, int col, char val, char[][] board){// 同行是否重复for (int i = 0; i < 9; i++){if (board[row][i] == val){return false;}}// 同列是否重复for (int j = 0; j < 9; j++){if (board[j][col] == val){return false;}}// 9宫格里是否重复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] == val){return false;}}}return true;}
}