[LeetCode-86] Partition List （鏈表數據分區）

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.

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【解題思路】

題意：
給定一個單鏈表和一個x，把鏈表中小於x的節點放到前面，
大於等於x的節點放到後面，每部分元素的原始相對位置不變。

思路：
其實很簡單，遍歷一遍鏈表，把小於x節點的都掛到small後，
把大於等於x的都放到big後面後，
最後再把大於等於的鏈表big掛到小於鏈表的後面就可以了。

【代碼實現】

struct ListNode* partition(struct ListNode* head, int x) 
{
	/*1.異常處理*/
	if(!head || !head->next) 
		return head;

	struct ListNode *big = (struct ListNode *)malloc(sizeof(struct ListNode));
	struct ListNode *small = (struct ListNode *)malloc(sizeof(struct ListNode));
	struct ListNode *b = big;
	struct ListNode *s = small;

	b->next = NULL;
	s->next = NULL;
	
	struct ListNode *p = head;
	
	struct ListNode *pnext = NULL;
	while (head) {
		pnext = head->next; //保存節點非常重要
		/*大於x的節點掛在x的後面，即掛在b的後面即可*/
		if(head->val >= x) { 
			b->next = head;
			b = b->next;
			b->next = NULL;
		/*小於x的節點掛在x的後面，小的掛在s的後面即可*/
		} else {
			s->next = head;
			s = s->next;
			s->next = NULL;
		}
		
		head = pnext;
	}

	s->next = big->next;/*大的結點掛在小的節點後面*/

	head = small->next;/*小的頭結點*/
	
	free(small);
	free(big);
	
	return head;
}