Linked Lists: Given a list, split it into two sublists — one for the fronthalf,

Question

Linked Lists:
Given a list, split it into two sublists — one for the fronthalf, and one for the back half. If the number of elements is odd,the extra element should go in the front list. So FrontBackSplit()on the list {2, 3, 5, 7, 11} should yield the two lists {2, 3, 5}and {7, 11}. Getting this right for all the cases is harder than itlooks. You should check your solution against a few cases (length =2, length = 3, length=4) to make sure that the list gets splitcorrectly near the short-list boundary conditions. If it worksright for length=4, it probably works right for length=1000. Youwill probably need special case code to deal with the (length<2) cases.

Hint: Probably the simplest strategy is to compute the length ofthe list, then use a for loop to hop over the right number of nodesto find the last node of the front halt, and then cut the list atthat point. there is a trick technique that uses two pointers totraverse the list. A "slow" pointer advances one nodes at a time,while the "fast" pointer goes two nodes at a time. when the fastpointer reaches the end, the slow pointer will be about half way.for either strategy, care is required to split the list at theright point. Linked Lists:
Given a list, split it into two sublists — one for the fronthalf, and one for the back half. If the number of elements is odd,the extra element should go in the front list. So FrontBackSplit()on the list {2, 3, 5, 7, 11} should yield the two lists {2, 3, 5}and {7, 11}. Getting this right for all the cases is harder than itlooks. You should check your solution against a few cases (length =2, length = 3, length=4) to make sure that the list gets splitcorrectly near the short-list boundary conditions. If it worksright for length=4, it probably works right for length=1000. Youwill probably need special case code to deal with the (length<2) cases.

Hint: Probably the simplest strategy is to compute the length ofthe list, then use a for loop to hop over the right number of nodesto find the last node of the front halt, and then cut the list atthat point. there is a trick technique that uses two pointers totraverse the list. A "slow" pointer advances one nodes at a time,while the "fast" pointer goes two nodes at a time. when the fastpointer reaches the end, the slow pointer will be about half way.for either strategy, care is required to split the list at theright point.

Explanation / Answer

Hi Nassin,
thanx alot for solving my Question

but there is an error..

fatal error C1083: cannotopen include file 'stdafx.h' No such file ordirectory

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