题目地址：

https://www.lintcode.com/problem/path-search/description

给定一个无向图，再给定两个点SSS和TTT，求所有SSS到TTT的简单路径，简单路径是指每个顶点最多只经过一次。要求路径按照字典序排序。

只需先建图，接着DFS的方法遍历一遍即可。注意，因为要按照字典序排序，所以建图的时候，邻居点要用TreeSet来存，这样在遍历路径的时候，字典序小的路径会先遍历。代码如下：

import java.util.*;public class Solution {/*** @param n: The number of points* @param G: The description of graph* @param S: The point S* @param T: The point T* @return: output all the paths from S to T*/public List<List<Integer>> getPath(int n, int[][] G, int S, int T) {// Write your code hereMap<Integer, Set<Integer>> graph = buildGraph(G);List<List<Integer>> res = new ArrayList<>();// cur表示当前路径，visited表示已经访问过的顶点。首先要把源点S加进去List<Integer> cur = new ArrayList<>();cur.add(S);Set<Integer> visited = new HashSet<>();visited.add(S);dfs(S, T, cur, res, visited, graph);return res;}private void dfs(int S, int T, List<Integer> cur, List<List<Integer>> res, Set<Integer> visited, Map<Integer, Set<Integer>> graph) {if (S == T) {res.add(new ArrayList<>(cur));return;}// 要验证一下在不在graph里，否则如果S没有邻居的话，就会造成NPEif (graph.containsKey(S)) {for (int neighbor : graph.get(S)) {if (!visited.contains(neighbor)) {cur.add(neighbor);visited.add(neighbor);dfs(neighbor, T, cur, res, visited, graph);// 回到上一层递归的时候要回溯，恢复现场visited.remove(neighbor);cur.remove(cur.size() - 1);}}}}private Map<Integer, Set<Integer>> buildGraph(int[][] G) {Map<Integer, Set<Integer>> graph = new HashMap<>();for (int[] edge : G) {// 邻居节点要用TreeSet存，这样遍历邻居的时候会从小到大遍历，路径也会按照字典序从小到大遍历graph.putIfAbsent(edge[0], new TreeSet<>());graph.putIfAbsent(edge[1], new TreeSet<>());graph.get(edge[0]).add(edge[1]);graph.get(edge[1]).add(edge[0]);}return graph;}
}

时空复杂度O(V+E)O(V+E)O(V+E)。

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【Lintcode】1647. Path Search

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