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完善java新方法的注释
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哈哈哈
@ -261,84 +261,6 @@ for (pair<const string, int>& target : targets[result[result.size() - 1]])
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### Java
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```java
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/* 首先遍历所有机票，将机票转换成map，其中起点作为key，终点按照字典顺序插入到终点的列表中
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然后进入递归，递归中首先在result中加入当前位置，如果票用完了就说明这条路走通了
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递归中遍历从当前位置出发的终点（也就是map中key为当前位置的列表），因为列表是有序的，因此只要出现能走通的路径，这条路径就是字典排序最低的
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在遍历中，首先从map中移除当前的机票，将遍历到的终点作为起点进入下一层的递归中，如果发现当前机票往后走走不通，再把当前的机票加到map中*/
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class Solution {
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//key为起点，value是有序的终点的列表
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Map<String, LinkedList<String>> ticketMap = new HashMap<>();
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LinkedList<String> result = new LinkedList<>();
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public List<String> findItinerary(List<List<String>> tickets) {
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//遍历tickets，存入ticketMap中
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for (List<String> ticket : tickets) {
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addNew(ticket.get(0), ticket.get(1));
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}
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deal("JFK");
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return result;
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}
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boolean deal(String currentLocation) {
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result.add(currentLocation);
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//机票全部用完，找到最小字符路径
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if (ticketMap.isEmpty()) {
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return true;
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}
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//当前位置的终点列表
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LinkedList<String> targetLocations = ticketMap.get(currentLocation);
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//机票没用完，但是没有从当前位置出发的机票了，说明这条路走不通
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if (targetLocations == null) {
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return false;
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}
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//终点列表中遍历到的终点
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String targetLocation;
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//遍历从当前位置出发的机票
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for (int i = 0; i < targetLocations.size(); i++) {
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targetLocation = targetLocations.get(i);
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//删除map中的机票，如果当前位置只有一个终点，直接删除k，v，有多个终点则删除终点列表中当前的终点
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if (targetLocations.size() == 1) {
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ticketMap.remove(currentLocation);
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} else {
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targetLocations.remove(i);
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}
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//递归
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if (deal(targetLocation)) {
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return true;
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} else {
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//路线走不通，将机票重新加到map中
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addNew(currentLocation, targetLocation);
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result.removeLast();
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}
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}
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return false;
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}
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/**
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* 在map中添加新元素
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*
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* @param start 起点
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* @param end 终点
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*/
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void addNew(String start, String end) {
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LinkedList<String> startAllEnd = ticketMap.get(start);
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if (startAllEnd != null) {
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for (int i = 0; i < startAllEnd.size() + 1; i++) {
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if (i == startAllEnd.size() || end.compareTo(startAllEnd.get(i)) < 0) {
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startAllEnd.add(i, end);
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break;
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}
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}
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} else {
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LinkedList<String> ends = new LinkedList<>();
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ends.add(end);
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ticketMap.put(start, ends);
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}
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}
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}
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```
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```java
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class Solution {
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private LinkedList<String> res;
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@ -423,6 +345,79 @@ class Solution {
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}
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```
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```java
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/* 该方法是对第二个方法的改进，主要变化在于将某点的所有终点变更为链表的形式，优点在于
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1.添加终点时直接在对应位置添加节点，避免了TreeMap增元素时的频繁调整
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2.同时每次对终点进行增加删除查找时直接通过下标操作，避免hashMap反复计算hash*/
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class Solution {
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//key为起点，value是有序的终点的列表
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Map<String, LinkedList<String>> ticketMap = new HashMap<>();
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LinkedList<String> result = new LinkedList<>();
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int total;
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public List<String> findItinerary(List<List<String>> tickets) {
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total = tickets.size() + 1;
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//遍历tickets，存入ticketMap中
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for (List<String> ticket : tickets) {
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addNew(ticket.get(0), ticket.get(1));
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}
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deal("JFK");
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return result;
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}
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boolean deal(String currentLocation) {
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result.add(currentLocation);
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//机票全部用完，找到最小字符路径
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if (result.size() == total) {
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return true;
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}
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//当前位置的终点列表
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LinkedList<String> targetLocations = ticketMap.get(currentLocation);
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//没有从当前位置出发的机票了，说明这条路走不通
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if (targetLocations != null && !targetLocations.isEmpty()) {
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//终点列表中遍历到的终点
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String targetLocation;
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//遍历从当前位置出发的机票
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for (int i = 0; i < targetLocations.size(); i++) {
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targetLocation = targetLocations.get(i);
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//删除终点列表中当前的终点
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targetLocations.remove(i);
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//递归
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if (deal(targetLocation)) {
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return true;
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}
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//路线走不通，将机票重新加回去
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targetLocations.add(i, targetLocation);
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result.removeLast();
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}
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}
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return false;
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}
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/**
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* 在map中按照字典顺序添加新元素
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*
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* @param start 起点
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* @param end 终点
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*/
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void addNew(String start, String end) {
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LinkedList<String> startAllEnd = ticketMap.getOrDefault(start, new LinkedList<>());
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if (!startAllEnd.isEmpty()) {
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for (int i = 0; i < startAllEnd.size(); i++) {
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if (end.compareTo(startAllEnd.get(i)) < 0) {
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startAllEnd.add(i, end);
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return;
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}
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}
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startAllEnd.add(startAllEnd.size(), end);
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} else {
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startAllEnd.add(end);
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ticketMap.put(start, startAllEnd);
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}
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}
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}
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```
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### Python
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回溯 使用used数组
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