【算法刷题day30】Leetcode：332. 重新安排行程、51. N 皇后、37. 解数独

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Leetcode 332. 重新安排行程

题目：332. 重新安排行程
解析：代码随想录解析

解题思路

代码

//第1版，通过 9/81（没看清题目，应该都是从JFK出发）
class Solution {List<String> res = new ArrayList<>();List<List<String>> paths = new ArrayList<>();boolean foundResult = false;boolean[] used;private void backtracking(List<List<String>> tickets) {if (foundResult)return;if (paths.size() == tickets.size()) {for (int i = 0; i < paths.size(); i++)res.add(paths.get(i).get(0));res.add(paths.get(paths.size()-1).get(1));foundResult = true;return;}for (int i = 0; i < tickets.size(); i++) {if (used[i])continue;if (!paths.isEmpty() && !(tickets.get(i).get(0)).equals(paths.get(paths.size() - 1).get(1)))continue;paths.add(tickets.get(i));used[i] = true;backtracking(tickets);paths.remove(paths.size()-1);used[i] = false;}}public List<String> findItinerary(List<List<String>> tickets) {//排序后找到的第一个。Collections.sort(tickets, new Comparator<List<String>>() {@Overridepublic int compare(List<String> o1, List<String> o2) {int setOff = o1.get(0).compareTo(o2.get(0));if (setOff != 0)return setOff;elsereturn o1.get(1).compareTo(o2.get(1));}});used = new boolean[tickets.size()];backtracking(tickets);return res;}
}//第2版，通过80/81，剩一个奇怪的样例没通过。和卡哥的第一份答案一样有个样例没通过
class Solution {List<String> res = new ArrayList<>();List<String> paths = new ArrayList<>();boolean foundResult = false;boolean[] used;private void backtracking(List<List<String>> tickets) {if (foundResult)return;if (paths.size() == tickets.size() + 1) {res.addAll(paths);foundResult = true;return;}for (int i = 0; i < tickets.size(); i++) {if (used[i])continue;if (!(tickets.get(i).get(0)).equals(paths.get(paths.size()-1)))continue;paths.add(tickets.get(i).get(1));used[i] = true;backtracking(tickets);paths.remove(paths.size()-1);used[i] = false;}}public List<String> findItinerary(List<List<String>> tickets) {//排序后找到的第一个。Collections.sort(tickets, new Comparator<List<String>>() {@Overridepublic int compare(List<String> o1, List<String> o2) {int setOff = o1.get(0).compareTo(o2.get(0));if (setOff != 0)return setOff;elsereturn o1.get(1).compareTo(o2.get(1));}});used = new boolean[tickets.size()];paths.add("JFK");backtracking(tickets);return res;}
}//第3版，通过81/81。使用 Map<String, Map<String, Integer>>来存储<出发站,<终点站, 票数>>，其中的终点站使用TreeMap来存储（保证了Key的有序性），比直接使用used更高效。
class Solution {List<String> res;Map<String, Map<String, Integer>> ticketMap;private boolean backtracking(int ticketNum) {if (res.size() == ticketNum + 1) {return true;}String destination = res.get(res.size()-1);if (ticketMap.containsKey(destination)) {for (Map.Entry<String, Integer> desNum : ticketMap.get(destination).entrySet()) {int count = desNum.getValue();if (count > 0) {res.add(desNum.getKey());desNum.setValue(count - 1);if (backtracking(ticketNum))return true;desNum.setValue(count);res.remove(res.size()-1);}}}return false;}public List<String> findItinerary(List<List<String>> tickets) {res = new ArrayList<>();ticketMap = new HashMap<>();for (List<String> ticket : tickets) {Map<String, Integer> tmp;if (ticketMap.containsKey(ticket.get(0))) {//是否有出发站的票tmp = ticketMap.get(ticket.get(0));//获取以此为出发站的终点站的票及对应票数量tmp.put(ticket.get(1), tmp.getOrDefault(ticket.get(1), 0) + 1);//让对应票数加1} else {tmp = new TreeMap<>();tmp.put(ticket.get(1), 1);}ticketMap.put(ticket.get(0), tmp);}res.add("JFK");backtracking(tickets.size());return res;}
}

总结

hard果然不容易

Leetcode 51. N 皇后

题目：51. N 皇后
解析：代码随想录解析

解题思路

每次是否下棋，就判断一下能否下这个皇后。一行一行下。一列一列判断。

代码

class Solution {List<List<String>> res;List<StringBuilder> chess;private boolean isValid(int row, int col, int n) {// for (int i = 0; i < n; i++)//     if (chess.get(row).charAt(i) == 'Q')//         return false;for (int i = 0; i < row; i++)if (chess.get(i).charAt(col) == 'Q')return false;for (int i = row-1, j = col-1; i >= 0 && j >= 0; i--, j--)if (chess.get(i).charAt(j) == 'Q')return false;for (int i = row-1, j = col+1; i >= 0 && j < n; i--, j++)if (chess.get(i).charAt(j) == 'Q')return false;return true;}private void backtracking(int n, int row) {if (row == n) {List<String> newChess = new ArrayList<>();for (int i = 0; i < chess.size(); i++)newChess.add(chess.get(i).toString());res.add(newChess);return;}for (int col = 0; col < n; col++) {if (isValid(row, col, n)) {chess.get(row).setCharAt(col, 'Q');backtracking(n, row + 1);chess.get(row).setCharAt(col, '.');}}}public List<List<String>> solveNQueens(int n) {res = new ArrayList<>();chess =  new ArrayList<>();for (int i = 0; i < n; i++) {StringBuilder sb = new StringBuilder();for (int j = 0; j < n; j++)sb.append('.');chess.add(sb);}backtracking(n, 0);return res;}
}

总结

暂无

Leetcode 37. 解数独

题目：37. 解数独
解析：代码随想录解析

解题思路

每次是否填入数字1-9，如果填入1-9都错误，则返回false。如果全部填完了，返回true。其中如果isValid已经达到了，就提前退出当前层的递归。

代码

class Solution {private boolean isValid(int row, int col, int val, char[][] board) {for (int i = 0; i < board.length; i++) {if (i == row)continue;if (board[i][col] == val)return false;}for (int j = 0; j < board.length; j++) {if (j == col)continue;if (board[row][j] == val)return false;}int beginRow = (row / 3) * 3;int beginCol = (col / 3) * 3;for (int i = beginRow; i < beginRow + 3; i++)for (int j = beginCol; j < beginCol + 3; j++) {if (i == row && j == col)continue;if (board[i][j] == val)return false;}return true;}private boolean backtracking(char[][] board) {for (int row = 0; row < board.length; row++) {for (int col = 0; col < board.length; col++) {if (board[row][col] == '.') {for (char val = '1'; val <= '9'; val++) {if (isValid(row, col, val, board)) {board[row][col] = val;if (backtracking(board))return true;board[row][col] = '.';}}return false;}}}return true;}public void solveSudoku(char[][] board) {backtracking(board);}
}

总结

暂无

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