feat: 标题分组

Author: sl1673495

算法/标题分组.js

@@ -0,0 +1,105 @@
+/**
+ * 需要输出
+    [{
+        "name": "h3"
+    }, {
+        "name": "h2",
+        "child": [{
+            "name": "h3"
+        }]
+    }, {
+        "name": "h1",
+        "child": [{
+            "name": "h2",
+            "child": [{
+                "name": "h3"
+            }, {
+                "name": "h3"
+            }]
+        }, {
+            "name": "h2",
+            "child": [{
+                "name": "h3"
+            }]
+        }]
+    }, {
+        "name": "h1",
+        "child": [{
+            "name": "h2",
+            "child": [{
+                "name": "h4"
+            }]
+        }, {
+            "name": "h2",
+            "child": [{
+                "name": "h3"
+            }]
+        }]
+    }]
+ */
+
+let list = [
+  'h3',
+  'h2', 'h3',
+  'h1', 'h2', 'h3', 'h3', 'h2', 'h3',
+  'h1', 'h2', 'h4', 'h2', 'h3'
+]
+
+function makeTree(arr) {
+  let tree = [];
+  let max = Infinity
+  let prev = {};
+
+  arr.forEach((title) => {
+    let level = Number(title[1]);
+    if (level <= max) {
+      let node = {
+        name: title,
+      };
+      tree.push(node);
+      prev.level = level;
+      prev.node = node;
+      max = level
+    } else if (level === prev.level) {
+      // 等级相同的话 上级节点的child继续增加子节点
+      prev = prev.prev;
+      pushNodeAndAdvanceQueue();
+    } else if (level > prev.level) {
+      pushNodeAndAdvanceQueue();
+    } else {
+      while (level <= prev.level) {
+        // 向上回溯，找到平级的再上一层node
+        prev = prev.prev;
+      }
+      pushNodeAndAdvanceQueue();
+    }
+
+    function pushNodeAndAdvanceQueue() {
+        let node = {
+            name: title,
+        };
+        if (!prev.node.child) {
+            let child = [];
+            prev.node.child = child;
+        }
+        prev.node.child.push(node);
+        let next = {
+          level,
+          node,
+          prev,
+        };
+        prev = next;
+    }
+  });
+
+  return tree;
+}
+
+console.log(makeTree(list));
+
+/**
+ * 思路
+ * 
+ * 利用链表来向上回溯合适的父节点
+ * 利用max变量记录当前分组的最大标题，如果比max还大，就需要新开一个分组了。
+ */