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
return
Solution :
The list can be traveled once to get the count of nodes and place a pointer at the last node.
In the next step, the list is traveled again from head, and each node having value greater than x is placed at the end of the list.
Code :
Please comment to suggest a better method.
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
.Solution :
The list can be traveled once to get the count of nodes and place a pointer at the last node.
In the next step, the list is traveled again from head, and each node having value greater than x is placed at the end of the list.
Code :
Please comment to suggest a better method.
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