先用使用常規方法，兩個指針:

golang實現：

type Node struct { value int next *Node}type Link struct { head *Node tail *Node lenth int}// 向鏈表中添加元素func (link *Link) add(v int) { if link.lenth == 0 { // 當前鏈表是空鏈表 link.head = &Node{v, nil} link.tail = link.head link.lenth = 1 } else { newNond := &Node{v, nil} link.tail.next = newNond link.tail = newNond link.lenth += 1 }}// 刪除鏈表中的元素(雙指針)func (link *Link) remove(v int) { if link.lenth == 0 { fmt.Println('空鏈表，不支持該操作') return } var previous *Node = nil for current := link.head; current != nil; current = current.next { if current.value == v { if current == link.head { // 要刪除的是頭節點link.head = current.next } else if current == link.tail { // 要刪除的是尾節點previous.next = nillink.tail = previous } else { // 要刪除的是中間的節點previous.next = current.next } link.lenth -= 1 break } previous = current }}// 打印鏈表func (link *Link) printList() { if link.lenth == 0 { fmt.Println('空鏈表') return } for cur := link.head; cur != nil; cur = cur.next { fmt.Printf('%d ', cur.value) } fmt.Println()}

python實現:

class Node: def __init__(self, value, next): self.value = value self.next = next def __str__(self): return str(self.value)class Link: def __init__(self): self.head = None self.tail = None self.lenth = 0 # 向鏈表中添加元素 def add(self, v): if self.lenth == 0: # 當前鏈表是空鏈表 self.head = Node(v, None) self.tail = self.head self.lenth = 1 else: new_node = Node(v, None) self.tail.next = new_node self.tail = new_node self.lenth += 1 # 打印鏈表 def print(self): if self.lenth == 0: print(’空鏈表’) return cur = self.head while True: if cur == None:print()break print(cur, end=’ ’) cur = cur.next # 刪除鏈表中的元素 def remove(self, v): if self.lenth == 0: return cur = self.head pre = None while True: if cur.value == v:if cur == self.head: # 要刪除的是頭節點 self.head = cur.nextelif cur == self.tail: # 要刪除的是尾節點 pre.next = None self.tail = preelse: # 要刪除的是中間的節點 pre.next = cur.nextself.lenth -= 1break pre = cur cur = cur.next if cur == None:print('未找到', v)break

只使用使用一個指針實現鏈表的刪除：

golang實現：

func (link *Link) remove_with_one_pointer(v int) { if link.lenth == 0 { return } if link.tail.value == v { // 要刪除的節點是尾節點，需特殊處理 if link.lenth == 1 { // 如果鏈表只有一個節點 link.head = nil link.tail = nil } else { //大于一個節點 cur := link.head for ; cur.next.next != nil; cur = cur.next { } //找到尾節點的前一個節點 cur.next = nil link.tail = cur } link.lenth -= 1 return } //要刪除的節點在頭部/中間 的常規情況 for cur := link.head; cur != nil; cur = cur.next { if cur.value == v { cur.value = cur.next.value cur.next = cur.next.next link.lenth -= 1 return } } fmt.Println('未找到', v)}

python實現：

def remove_with_one_pointer(self, v): if self.lenth == 0: return if self.tail.value == v: # 要刪除的節點是尾節點，需特殊處理 if self.lenth == 1: # 如果鏈表只有一個節點 self.head = None self.tail = None else: # 大于一個節點 cur = self.head while True:if cur.next.next is None: # 找到尾節點的前一個節點 breakelse: cur = cur.next cur.next = None self.tail = cur self.lenth -= 1 return # 要刪除的節點在頭部/中間 的常規情況 cur = self.head while True: if cur.value == v: cur.value = cur.next.value cur.next = cur.next.next self.lenth -= 1 break cur = cur.next if cur is None: print(’未找到’, v) break

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