LeetCode 19. Remove Nth Node From End of List

链表题。很简单但是debug了非常久。

愚蠢如我...把所有地址存在数组里。空间复杂度...

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
        ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode *p=head,*q;
        ListNode *s[100000];
        int m=0;
        while(p){
            s[m]=p;m++;
            p=p->next;
        }
        if (m==n) return head->next;
        if(n==1) (*s[m-2]).next=NULL;
        else {
            p=s[m-n-1];
            q=p->next;
            (*p).next=(*q).next;
        }
        return head;
    }
};

 

非常巧妙的方法，保持两个指针差n个位置，使后面的指针指向最后一个，前面的指针即为倒数第n.

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(!head->next) return NULL;
        ListNode *pre = head;
        ListNode *cur = head;
        for(int i = 0;i<n;i++) cur = cur->next;
        if(!cur) return head->next;
        while(cur->next){
            cur = cur->next;
            pre = pre->next;
        }
        pre->next = pre->next->next;
        return head;
    }
};