【学习打卡】第26天 数据结构和算法

堆

合并K个升序链表（leetcode - 23）

给你一个链表数组，每个链表都已经按升序排列。
请你将所有链表合并到一个升序链表中，返回合并后的链表。

示例

输入：lists = [[1,4,5],[1,3,4],[2,6]]
输出：[1,1,2,3,4,4,5,6]
解释：链表数组如下：
[
  1->4->5,
  1->3->4,
  2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6

思路

1. 创建一个最小堆，将链表数组的每个链表头部插入堆中
2. 循环最小堆，将堆顶出堆，将堆顶的下一个链表插入堆中
3. 堆为空时循环结束，返回指针的头部

class MinHeap {
  constructor() {
    this.heap = [];
  }

  swap(i1, i2) {
    const temp = this.heap[i1];
    this.heap[i1] = this.heap[i2];
    this.heap[i2] = temp;
  }

  getParentIndex(index) {
    return Math.floor((index - 1) / 2);
  }

  getLeftIndex(index) {
    return index * 2 + 1;
  }

  getRightIndex(index) {
    return index * 2 + 2;
  }

  shiftUp(index) {
    if (index === 0) return
    const parentIndex = this.getParentIndex(index);
    if (this.heap[parentIndex] && this.heap[parentIndex].val > this.heap[index].val) {
      this.swap(index, parentIndex);
      this.shiftUp(parentIndex)
    }
  }

  shiftDown(index) {
    if (index === this.heap.length - 1) return;
    const leftIndex = this.getLeftIndex(index);
    const rightIndex = this.getRightIndex(index);
    if (this.heap[leftIndex] &&  this.heap[leftIndex].val < this.heap[index].val) {
      this.swap(index, leftIndex);
      this.shiftDown(leftIndex);
    }
    if (this.heap[rightIndex] && this.heap[rightIndex].val < this.heap[index].val) {
      this.swap(index, rightIndex);
      this.shiftDown(rightIndex);
    }
  }

  insert(val) {
    this.heap.push(val);
    this.shiftUp(this.heap.length - 1);
  }

  pop() {
    if(this.size() === 1) return this.heap.shift();
    const top = this.heap[0];
    this.heap[0] = this.heap.pop();
    this.shiftDown(0);
    return top;
  }

  peek() {
    return this.heap[0];
  }

  size() {
    return this.heap.length;
  }
}
/**
 * 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}
 */
var mergeKLists = function(lists) {
    const res = new ListNode(0);
    const h = new MinHeap();
    let p = res;
    lists.forEach(l => {
        if(l) h.insert(l)
    });
    while(h.size()) {
        const n = h.pop();
        p.next = n;
        p = p.next;
        if(n.next) h.insert(n.next);
    }
    return res.next;
};

复杂度分析：
时间复杂度：O(nlogk)
空间复杂度：O(k)