本文主要是介绍leetcode|剑指 Offer 22. 链表中倒数第k个节点,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
class Solution {
public:
ListNode* getKthFromEnd(ListNode* head, int k) {
int myCount = 0;
ListNode* fast = head;
ListNode* slow = head;
if(!head){
return head;
}
while(fast->next && fast->next->next){
myCount++;
fast = fast->next->next;
slow = slow->next;
}
int size = 0;
if(fast->next==NULL){
size = myCount*2+1;
}else{
size = (myCount+1)*2;
}
int index = size-k;
if(index>=myCount){//需要的节点在中间节点(慢指针)以后
ListNode* res = slow;
for(int i=index-myCount;i>0;i--){
res = res->next;
}
return res;
}else{
ListNode* res = head;
for(int i=0;i<index;i++){
res = res->next;
}
return res;
}
}
};
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