✨ reconstruct itinerary

Author: tangweikun

src/322-reconstruct-itinerary/index.js

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+/**
+ * @param {string[][]} tickets
+ * @return {string[]}
+ */
+// HELP:
+var findItinerary = function (tickets) {
+  const graph = getGraph(tickets);
+  const ans = [];
+  dfs(graph, 'JFK', ans);
+  // 深度优先遍历是先进先出，所以reverse反转一下
+  return ans.reverse();
+};
+
+//从JFS出发，深度优先遍历
+function dfs(graph, cur, ans) {
+  if (!graph.has(cur)) return;
+  const neighbors = graph.get(cur);
+  // 题目要求先遍历字典序小的元素
+  while (neighbors.length) dfs(graph, neighbors.shift(), ans);
+  // 因为深度优先遍历是先进先出，所以每次遍历出发点添加到最后，意为最先出
+  ans.push(cur);
+}
+
+// 先存储所有机场之间的关系，哪怕tickets中没有给某个机场提供目的地记录，也要把它的目的地置为空数组
+function getGraph(tickets) {
+  const map = new Map();
+  // 收集每一个出发点的所有目的地
+  for (let i = 0; i < tickets.length; i++) {
+    const from = tickets[i][0];
+    const to = tickets[i][1];
+    if (!map.has(from)) map.set(from, []);
+    if (!map.has(to)) map.set(to, []);
+    map.get(from).push(to);
+  }
+  for (let [key, value] of map) {
+    // 字典顺序排序目的地数组
+    value.sort((a, b) => (a < b ? -1 : a > b ? 1 : 0));
+  }
+
+  return map;
+}

src/322-reconstruct-itinerary/pro.js

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+/**
+ * @param {string[][]} tickets
+ * @return {string[]}
+ */
+var findItinerary = function (tickets) {
+  let map = {},
+    result = [];
+  for (let i of tickets) {
+    if (!map[i[0]]) {
+      map[i[0]] = [i[1]];
+    } else {
+      map[i[0]].push(i[1]);
+    }
+    if (!map[i[1]]) {
+      map[i[1]] = [];
+    }
+  }
+
+  getResult(map, 'JFK', result);
+  function getResult(map, spot, result) {
+    //   console.log(map)
+    if (!map[spot]) return;
+    const neighbors = map[spot];
+    neighbors.sort();
+    while (neighbors.length > 0) {
+      getResult(map, neighbors.shift(), result);
+    }
+    result.push(spot);
+  }
+
+  return result.reverse();
+};