[LeetCode刷题记录]Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 1->4->3->2->5->2 and x = 3,

return 1->2->2->4->3->5.

思路：用两个ListNode一个来链接原来List中比x小的，一个用来链接原来List中比x大的。最后把两个List链接起来，删掉多余节点。

空间O(1)时间O(n)。

class Solution {
public:
	ListNode *partition(ListNode *head, int x) {
		if(!head || !head->next) return head;
		ListNode *big = new ListNode(1);
		ListNode *small = new ListNode(-1);
		ListNode *b = big;
		ListNode *s = small;
		ListNode *p = head;
		while (head)
		{
			ListNode *pnext = head->next;
			if(head->val >= x){
				head->next = NULL;
				b->next = head;
				b = head;
			}
			else{
				head->next = NULL;
				s->next = head;
				s = head;
			}
			head = pnext;
		}
		s->next = big->next;
		head = small->next;
		delete small;
		delete big;
		return head;
	}
};

如果想保留原来List的话：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        ListNode *small = new ListNode(0);
        ListNode *large = new ListNode(0);
        ListNode *pre = head;
        ListNode *ret = small;
        ListNode *temp = large;
        if(head==NULL)
            return NULL;
        while(pre!=NULL){
            if(pre->val<x){
                small->next = new ListNode(pre->val);
                small = small->next;
            }
            else{
                large->next = new ListNode(pre->val);
                large = large->next;
            }
            pre=pre->next;
        }
        small->next = temp->next;
        delete temp;
        return ret->next;
    }
};