1: /** 2: * Definition for singly-linked list. 3: * struct ListNode { 4: * int val; 5: * ListNode *next; 6: * ListNode(int x) : val(x), next(NULL) {} 7: * }; 8: */ 9: class Solution { 10: public: 11: ListNode* partition(ListNode* head, int x) { 12: ListNode *dummy = new ListNode(-1); 13: ListNode *prev = dummy, *cur = head, *par = dummy; 14: dummy->next = head; 15: while (cur) { 16: if (cur->val < x) { 17:
if (prev == par) {
18: cur = cur->next; 19: prev = prev->next; 20: par = par->next; 21: } else { 22: prev->next = cur->next; 23: cur->next = par->next; 24: par->next = cur; 25: cur = prev->next; 26: par = par->next; 27: } 28: } else { 29: prev = cur; 30: cur = cur->next; 31: } 32: } 33: return dummy->next; 34: } 35: };
Another solution will be separate the list into two lists and combine them in the end.
When I revisited this problem, I had a bit more concise solution.
1: /**
2: * Definition for singly-linked list.
3: * struct ListNode {
4: * int val;
5: * ListNode *next;
6: * ListNode(int x) : val(x), next(NULL) {}
7: * };
8: */
9: class Solution {
10: public:
11: ListNode* partition(ListNode* head, int x) {
12: ListNode *dummy = new ListNode(-1);
13: ListNode *small = dummy;
14: ListNode *large = head, *large_head = NULL, *large_prev = NULL;
15: while (large) {
16: if (large->val < x) {
17: small->next = large;
18: large = large->next;
19: if (large_prev) large_prev->next = large;
20: small = small->next;
21: small->next = NULL;
22: } else {
23: if (large_head == NULL) large_head = large;
24: large_prev = large;
25: large = large->next;
26: }
27: }
28: small->next = large_head;
29: head = dummy->next;
30: delete dummy;
31: return head;
32: }
33: };
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