From 0a41e81e588e7d2a38552f42e1bca078a501ad2d Mon Sep 17 00:00:00 2001
From: LouisFonda <yigencong@yahoo.com>
Date: Sat, 7 Jun 2025 16:08:53 +0800
Subject: [PATCH] =?UTF-8?q?feat:=20=E5=88=86=E6=B2=BB=E7=AE=97=E6=B3=9523,?=
 =?UTF-8?q?10,148,427?=
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Content-Transfer-Encoding: 8bit

---
 .../divide/108将有序数组转换为二叉树搜索树.js |  38 ++++
 .../divide/148排序列表.js                     | 166 ++++++++++++++++++
 .../divide/23合并k个升序列表.js               | 112 ++++++++++++
 .../divide/427构建四叉树.js                   |  93 ++++++++++
 4 files changed, 409 insertions(+)
 create mode 100644 top-interview-leetcode150/divide/108将有序数组转换为二叉树搜索树.js
 create mode 100644 top-interview-leetcode150/divide/148排序列表.js
 create mode 100644 top-interview-leetcode150/divide/23合并k个升序列表.js
 create mode 100644 top-interview-leetcode150/divide/427构建四叉树.js

diff --git a/top-interview-leetcode150/divide/108将有序数组转换为二叉树搜索树.js b/top-interview-leetcode150/divide/108将有序数组转换为二叉树搜索树.js
new file mode 100644
index 0000000..c8cc494
--- /dev/null
+++ b/top-interview-leetcode150/divide/108将有序数组转换为二叉树搜索树.js
@@ -0,0 +1,38 @@
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {number[]} nums
+ * @return {TreeNode}
+ */
+const sortedArrayToBST = function (nums) {
+
+};
+
+function TreeNode(val, left, right) {
+  this.val = (val === undefined ? 0 : val);
+  this.left = (left === undefined ? null : left);
+  this.right = (right === undefined ? null : right);
+}
+
+/*
+用数组的中间值作为根节点，然后递归的将左半部分和右半部分处理成一颗二叉搜索树，并设置成根节点的左右子树
+*/
+function f1(left, right, nums) {
+  // 如果left > right, 说明这个区间没有元素，直接返回null
+  if (left > right) return null;
+  // 计算中间节点的索引
+  const mid = left + Math.floor((right - left) / 2);
+
+  // 创建根节点，或子树的根节点
+  const root = new TreeNode(nums[mid]);
+  // 递归处理左半部分和右半部分
+  root.left = f1(left, mid - 1, nums);
+  root.right = f1(mid + 1, right, nums);
+  return root;
+}
diff --git a/top-interview-leetcode150/divide/148排序列表.js b/top-interview-leetcode150/divide/148排序列表.js
new file mode 100644
index 0000000..36d5504
--- /dev/null
+++ b/top-interview-leetcode150/divide/148排序列表.js
@@ -0,0 +1,166 @@
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+
+function ListNode(val, next) {
+  this.val = (val === undefined ? 0 : val);
+  this.next = (next === undefined ? null : next);
+}
+
+/**
+ * @param {ListNode} head
+ * @return {ListNode}
+ */
+const sortList = function (head) {
+
+};
+
+/*
+利用归并排序的思想，将链表分成两半，分别对两半进行排序，然后合并两个已排序的链表。
+*/
+function f1(head) {
+  if (!head || !head.next) return head;
+  let tail = head;
+  while (tail.next) {
+    tail = tail.next;
+  }
+
+  return merge(head, tail);
+}
+
+/*
+将一个链表归并成一个有序列表，具体过程可以参考归并排序，不好描述，但不是很难
+*/
+function merge(head, tail) {
+  if (head === tail) return head;
+
+  // 快慢指针计算中间节点
+  let low = head;
+  let fast = head;
+
+  while (fast.next && fast.next.next) {
+    low = low.next;
+    fast = fast.next.next;
+  }
+
+  const mid = low;
+  // 将链表从中间截断
+  const head2 = mid.next;
+  mid.next = null;
+
+  // 将截断后的两个链表继续进行merge操作，之后将其合并成一个有序的链表返回
+  return mergeList(merge(head, mid), merge(head2, tail));
+}
+
+/*
+  合并两个有序链表
+ */
+function mergeList(l1, l2) {
+  // 至少一个为空的情况
+  if (!l1 || !l2) {
+    // 返回其中任意一个不为空的情况
+    return l1 || l2;
+  }
+
+  // 两个链表都不为空，将两个链表中的所有节点按大小拼接到dummy上，最后返回dummy.next
+  const dummy = new ListNode(0);
+  let cur = dummy;
+  while (l1 && l2) {
+    if (l1.val <= l2.val) {
+      cur.next = l1;
+      l1 = l1.next;
+    } else {
+      cur.next = l2;
+      l2 = l2.next;
+    }
+    cur = cur.next;
+  }
+  // 当一个链表处理完毕，还会右一个列表存在节点，将它拼接到cur.next上
+  cur.next = l1 === null ? l2 : l1;
+  return dummy.next;
+}
+
+/*
+优化上面的写法，思路一致，但是只传入一个参数head，表示一个链表的头节点
+*/
+function f2(head) {
+  // 如果链表为空，或者只有一个节点，直接返回head
+  if (head === null || head.next === null) {
+    return head;
+  }
+
+  // 通过快慢指针寻找中间节点，将链表分成左右两部分
+  let slow = head;
+  let fast = head;
+  let prev = null;
+  while (fast && fast.next) {
+    prev = slow;
+    slow = slow.next;
+    fast = fast.next.next;
+  }
+
+  // 把链表从中间断开
+  prev.next = null;
+
+  const left = f2(head);
+  const right = f2(slow);
+  return mergeList(left, right);
+}
+
+/*
+利用快速排序的思想，将链表分成三部分，left, pivot, right,left表示小于pivot的部分，right表示大于pivot的部分,最后将它们拼接起来。
+*/
+function quickSortList(head) {
+  if (!head || !head.next) return head;
+
+  const pivot = head;
+  const leftDummy = new ListNode(0);
+  const rightDummy = new ListNode(0);
+  let leftCur = leftDummy;
+  let rightCur = rightDummy;
+  let cur = head.next;
+
+  // 分组：< pivot 到 left，≥ pivot 到 right
+  while (cur) {
+    if (cur.val < pivot.val) {
+      leftCur.next = cur;
+      leftCur = leftCur.next;
+    } else {
+      rightCur.next = cur;
+      rightCur = rightCur.next;
+    }
+    cur = cur.next;
+  }
+
+  // 截断，避免串联成环
+  leftCur.next = null;
+  rightCur.next = null;
+  pivot.next = null; // 断开 pivot 和旧链表的联系
+
+  const leftSorted = quickSortList(leftDummy.next);
+  const rightSorted = quickSortList(rightDummy.next);
+
+  // 拼接三段：leftSorted + pivot + rightSorted
+  return concat(leftSorted, pivot, rightSorted);
+}
+
+// 拼接：将 left + pivot + right 接成一个链表并返回头结点
+function concat(left, pivot, right) {
+  let head = pivot;
+
+  if (left) {
+    head = left;
+    let tail = left;
+    while (tail.next) {
+      tail = tail.next;
+    }
+    tail.next = pivot;
+  }
+
+  pivot.next = right;
+  return head;
+}
diff --git a/top-interview-leetcode150/divide/23合并k个升序列表.js b/top-interview-leetcode150/divide/23合并k个升序列表.js
new file mode 100644
index 0000000..4b4d3c3
--- /dev/null
+++ b/top-interview-leetcode150/divide/23合并k个升序列表.js
@@ -0,0 +1,112 @@
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode[]} lists
+ * @return {ListNode}
+ */
+const mergeKLists = function (lists) {
+
+};
+
+function ListNode(val, next) {
+  this.val = (val === undefined ? 0 : val);
+  this.next = (next === undefined ? null : next);
+}
+
+/*
+定义一个minIndex变量来记录当前最小头节点的索引。
+在每次循环中，遍历lists数组，找到当前最小头节点的索引minIndex。
+如果找到的minIndex为-1，说明没有更多的节点可以合并，跳出循环。
+如果找到的minIndex不为-1，将当前最小头节点添加到合并后的链表中。
+然后将lists[minIndex]指向下一个节点，继续下一轮循环。
+*/
+
+/**
+ * 合并k个升序链表
+ * @param {ListNode[]} lists
+ * @return {ListNode}
+ */
+function f1(lists) {
+  const dummy = new ListNode(0);
+  let current = dummy;
+
+  // 遍历lists，找到最小的头节点
+  while (true) {
+    let minIndex = -1;
+
+    for (let i = 0; i < lists.length; i++) {
+      if (lists[i] && (minIndex === -1 || lists[i].val < lists[minIndex].val)) {
+        minIndex = i; // 更新最小头节点的索引
+      }
+    }
+
+    if (minIndex === -1) break; // 没有找到最小的头节点
+
+    current.next = lists[minIndex]; // 拼接到dummy节点上
+    current = current.next; // 移动到下一个节点
+    lists[minIndex] = lists[minIndex].next; // 移动到下一个头节点
+  }
+  return dummy.next; // 返回合并后的链表
+}
+
+/*
+定义一个ans表示要返回的链表，然后将lists中的链表挨个和它合并，最终返回ans
+ */
+function f2(lists) {
+  let ans = null;
+  for (let i = 0; i < lists.length; i++) {
+    ans = mergeLists(ans, lists[i]);
+  }
+  return ans;
+}
+
+/*
+使用分治法
+ */
+function f3(lists) {
+  return merge(lists, 0, lists.length - 1);
+}
+
+/**
+*  @param {ListNode[]} lists 链表数组
+*   @param {number} l 要进行合并的左边界
+*   @param {number} r 要进行合并的右边界
+* merge函数会将l到r范围的链表合并成一个新链表
+ */
+function merge(lists, l, r) {
+  // 如果l和r相等，表明这个范围只有一个链表，直接返回
+  if (l === r) return lists[l];
+  if (l > r) return null;
+  // 将l到r的范围拆分成左右两部分，等这两部分merge之后再合并它们
+  const mid = Math.floor((l + r) / 2);
+
+  return mergeLists(merge(lists, l, mid), merge(lists, mid + 1, r));
+}
+
+function mergeLists(l1, l2) {
+  if (l1 === null || l2 === null) {
+    return l1 === null ? l2 : l1;
+  }
+
+  const dummy = new ListNode(0);
+  let cur = dummy;
+
+  while (l1 && l2) {
+    if (l1.val <= l2.val) {
+      cur.next = l1;
+      l1 = l1.next;
+    } else {
+      cur.next = l2;
+      l2 = l2.next;
+    }
+    cur = cur.next; // 移动到下一个节点
+  }
+
+  cur.next = l1 === null ? l2 : l1; // 将剩余的节点拼接到链表后面
+  return dummy.next; // 返回合并后的链表
+}
diff --git a/top-interview-leetcode150/divide/427构建四叉树.js b/top-interview-leetcode150/divide/427构建四叉树.js
new file mode 100644
index 0000000..e4be595
--- /dev/null
+++ b/top-interview-leetcode150/divide/427构建四叉树.js
@@ -0,0 +1,93 @@
+/* eslint-disable max-len */
+// eslint-disable-next-line no-underscore-dangle
+function _Node(val, isLeaf, topLeft, topRight, bottomLeft, bottomRight) {
+  this.val = val;
+  this.isLeaf = isLeaf;
+  this.topLeft = topLeft;
+  this.topRight = topRight;
+  this.bottomLeft = bottomLeft;
+  this.bottomRight = bottomRight;
+}
+
+/**
+ * @param {number[][]} grid
+ * @return {_Node}
+ */
+const construct = function (grid) {
+
+};
+
+/*
+利用分治算法，我们很容易发现当啊n=1时grid一定可以构建成一个叶子节点，val=当前这个位置表示的值，当n=2时也即是由四个n=1的grid组成的，
+我们把它们划分好，然后查看这左上，有伤，左下，右下的这四个四叉树节点的值是否都是叶子节点，如果都是再判断它们的值是否相等，如果是叶子节点
+并且它们的值也相等，就把他们合成一个更大的叶子节点，否则返回这个有叶子节点的四叉树
+*/
+function f1(grid) {
+  const n = grid.length;
+  return buildNode(grid, 0, n - 1, 0, n - 1);
+}
+
+/**
+ *
+ * @param {*} grid
+ * @param {*} start
+ * @param {*} end
+ */
+function buildNode(grid, rStart, rEnd, cStart, cEnd) {
+  // 如果划分的grid只有一个值，那么这个值就是四叉树的叶子节点
+  if (rStart === rEnd) {
+    return new _Node(grid[rStart][cStart], 1);
+  }
+
+  // 划分四个区域构建子节点
+  const topLeft = buildNode(grid, rStart, rStart + Math.floor((rEnd - rStart) / 2), cStart, cStart + Math.floor((cEnd - cStart) / 2));
+  const topRight = buildNode(grid, rStart, rStart + Math.floor((rEnd - rStart) / 2), cStart + Math.floor((cEnd - cStart) / 2) + 1, cEnd);
+  const bottomLeft = buildNode(grid, rStart + Math.floor((rEnd - rStart) / 2) + 1, rEnd, cStart, cStart + Math.floor((cEnd - cStart) / 2));
+  const bottomRight = buildNode(grid, rStart + Math.floor((rEnd - rStart) / 2) + 1, rEnd, cStart + Math.floor((cEnd - cStart) / 2) + 1, cEnd);
+
+  // 如果四个节点都是叶子节点，并且值相同将它们合并成一个更大的叶子节点
+  if (topLeft.isLeaf && topRight.isLeaf && bottomLeft.isLeaf && bottomRight.isLeaf && topLeft.val === topRight.val && topRight.val === bottomLeft.val && bottomLeft.val === bottomRight.val) {
+    return new _Node(topLeft.val, 1);
+  }
+
+  // 构建一个包含四个节点的非叶子节点
+  return new _Node(1, 0, topLeft, topRight, bottomLeft, bottomRight);
+}
+
+/*
+优化分治思路
+*/
+function buildNodeBest(grid) {
+  const n = grid.length;
+
+  function dfs(r0, c0, size) {
+    // 判断是否整个区域都是同一个值
+    const val = grid[r0][c0];
+    let same = true;
+    for (let i = r0; i < r0 + size; i++) {
+      for (let j = c0; j < c0 + size; j++) {
+        if (grid[i][j] !== val) {
+          same = false;
+          break;
+        }
+      }
+      if (!same) break;
+    }
+
+    if (same) {
+      return new Node(val === 1, true, null, null, null, null);
+    }
+
+    const half = size / 2;
+    return new Node(
+      true,
+      false,
+      dfs(r0, c0, half), // topLeft
+      dfs(r0, c0 + half, half), // topRight
+      dfs(r0 + half, c0, half), // bottomLeft
+      dfs(r0 + half, c0 + half, half), // bottomRight
+    );
+  }
+
+  return dfs(0, 0, n);
+}