题解 | 链表中倒数最后k个结点
链表中倒数最后k个结点
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/** * struct ListNode { * int val; * struct ListNode *next; * ListNode(int x) : val(x), next(nullptr) {} * }; */ class Solution { public: /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param pHead ListNode类 * @param k int整型 * @return ListNode类 */ ListNode* FindKthToTail(ListNode* pHead, int k) { // write code here //1.递归,从顶到底后从底到顶找到倒数第k个节点; //2.遍历后知道长度,可用栈\向量存储遍历结点,之后遍历k或者索引到目标节点; //3.快慢指针(推荐):https://blog.nowcoder.net/n/d82ed71f5481490ca79c4f0cd3103749; // ListNode* pBody=NULL; recursion(pHead, k, pBody); if(k!=0){ return NULL; } return pBody; } void recursion(ListNode* pHead, int& k, ListNode*& pBody){ if(pHead ==NULL){ return; } else{ recursion(pHead->next, k, pBody); if(k>0){ pBody = pHead; k--; } return; } } };