LeetCode:23.合并K个排序链表
解题思路新链表的下一个节点一定是k个链表头中的最小节点。考虑选择使用最小堆。
解题步骤构建一个最小堆，并依次把链表头插入堆中。弹出堆顶接到输出链表，并将堆顶所在链表的新链表头插入堆中。等堆元素全部弹出，合并工作就完成了。
class MinHeap {constructor() {this.heap = []this.len = 0}size() {return this.len}push(val) {this.heap[++this.len] = valthis.swin(this.len)}pop() {if(this.top()===1) return this.heap.shift(1);const ret = this.heap[1]this.swap(1, this.len--)this.heap[this.len + 1] = nullthis.sink(1)return ret}swin(ind) {while (ind > 1 && this.less(ind, parseInt(ind / 2))) {this.swap(ind, parseInt(ind / 2))ind = parseInt(ind / 2)}}sink(ind) {while (ind * 2 <= this.len) {let j = ind * 2if (j < this.len && this.less(j + 1, j)) j++if (this.less(ind, j)) breakthis.swap(ind, j)ind = j}}top() {return this.heap[1]}isEmpty() {return this.len === 0}swap(i, j) {[this.heap[i], this.heap[j]] = [this.heap[j], this.heap[i]]}less(i, j) {return this.heap[i].val < this.heap[j].val}getHeap() {return this.heap}}
/*** 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) {if(!lists) return;let root=new ListNode(0);let h=new MinHeap()lists.forEach(n=>n&&h.push(n))let n;let p=rootwhile(h.size()){n=h.pop();p.next=np=p.nextif(n.next)h.push(n.next)}return root.next
};'
