Week 41 @ 2024 算法周记

Merge Two Sorted Lists

https://leetcode.com/problems/merge-two-sorted-lists/

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeTwoLists(list1 *ListNode, list2 *ListNode) *ListNode {
    dummy := &ListNode{}
    p := dummy
    
    p1 := list1
    p2 := list2
    
    for p1 != nil && p2 != nil {
        if p1.Val < p2.Val {
            p.Next = p1
            p1 = p1.Next
            p = p.Next
        } else {
            p.Next = p2
            p2 = p2.Next
            p = p.Next
        }
    }
    
    for p1 != nil {
        p.Next = p1
        p1 = p1.Next
        p = p.Next
    }
    
    for p2 != nil {
        p.Next = p2
        p2 = p2.Next
        p = p.Next
    }
    
    return dummy.Next
}

优化解法

最后去合并剩余的 list1 / list2 时，不需要逐节点合并，只需要让 p.Next 指向最终剩余的 list 即可（其他节点已经「组成好一个有序链表」了）

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeTwoLists(list1 *ListNode, list2 *ListNode) *ListNode {
    dummy := &ListNode{}
    p := dummy
    
    p1 := list1
    p2 := list2
    
    for p1 != nil && p2 != nil {
        if p1.Val < p2.Val {
            p.Next = p1
            p1 = p1.Next
            p = p.Next
        } else {
            p.Next = p2
            p2 = p2.Next
            p = p.Next
        }
    }
    
    if p1 != nil {
        p.Next = p1
    }
    
    if p2 != nil {
        p.Next = p2
    }
    
    return dummy.Next
}

Partition List

https://leetcode.com/problems/partition-list/

题意：看上去返回的是一个链表，实际上是返回了两个链表连起来的。。

相当于生成两个新的链表，一个链表的所有值都 < x，另一个都 >=x ，再将这两个链表合到一起

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func partition(head *ListNode, x int) *ListNode {
    dummy1 := &ListNode{}
    dummy2 := &ListNode{}
    p1 := dummy1
    p2 := dummy2
    
    
    for p := head; p != nil; p = p.Next {
        if p.Val < x {
            p1.Next = p
            p1 = p1.Next
        } else {
            p2.Next = p
            p2 = p2.Next
        }
    }
    
    p1.Next = dummy2.Next
    p2.Next = nil
    
    return dummy1.Next
}

Merge k Sorted Lists

https://leetcode.com/problems/merge-k-sorted-lists/

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeKLists(lists []*ListNode) *ListNode {
    dummy := &ListNode{}
    p := dummy
    
    for {
        hasAny := false
        
        mi := -1
        mv := 1000000
        mp := (*ListNode)(nil)
        
        for i, h := range lists {
            if h == nil {
                continue
            }
            hasAny = true
            
            if h.Val < mv {
                mi = i
                mv = h.Val
                mp = h
            }
        }
        
        if !hasAny {
            break
        }
        
        p.Next = mp
        p = p.Next
        lists[mi] = lists[mi].Next
    }
    
    return dummy.Next
}

上述代码时间复杂度为 O(lists.length * max(lists[i].length))，即 O(NM)

考虑条件

0 <= lists[i].length <= 500
k == lists.length, 0 <= k <= 10^4

最大为 5 * 10^6 可以通过 OJ

优化方案

将所有优化方案

将所有 list 的头节点加入堆中，时间复杂度降为 O(sum(lists[i].length) * log(lists.length))，即 O(NlogK)

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeKLists(lists []*ListNode) *ListNode {
    dummy := &ListNode{}
    p := dummy
    
    pq := &NodeHeap{}
    for _, l := range lists {
        if l != nil {
            heap.Push(pq, l)
        }
    }
    
    for pq.Len() != 0 {
        v := heap.Pop(pq).(*ListNode)
        p.Next = v
        p = p.Next
        
        if v.Next != nil {
            heap.Push(pq, v.Next)
        }
    }
    
    return dummy.Next
}

type NodeHeap []*ListNode

func (h NodeHeap) Len() int {
    return len(h)
}
func (h NodeHeap) Swap(i, j int) {
    h[i], h[j] = h[j], h[i]
}
func (h NodeHeap) Less(i, j int) bool {
    return h[i].Val < h[j].Val
}

func (h *NodeHeap) Push(x any) {
    *h = append(*h, x.(*ListNode))
}
func (h *NodeHeap) Pop() any {
    n := len(*h)
    
    v := (*h)[n-1]
    *h = (*h)[:n-1]
    
    return v
}