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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.
/** * 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) { if(head == NULL) return NULL; ListNode *small, *big, *shead, *bhead; shead = (ListNode *)malloc(sizeof(ListNode)); bhead = (ListNode *)malloc(sizeof(ListNode)); shead->next = bhead->next = NULL; small = shead; big = bhead; while(head != NULL){ if(head->val >= x){ ListNode *tmp = (ListNode *)malloc(sizeof(ListNode)); tmp->val = head->val; tmp->next = NULL; big->next = tmp; big = tmp; } else { ListNode *tmp = (ListNode *)malloc(sizeof(ListNode)); tmp->val = head->val; tmp->next = NULL; small->next = tmp; small = tmp; } head = head->next; } small->next = bhead->next; return shead->next; }};
Partition List
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