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                Sorting singly linked list which takes only one argumentover 9 years ago 
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          about 9 years ago I have a solution, not the fastest one, but it respect the conditions. Imagine your list is : class LinkedInt { int integer; LinkedInt next; LinkedInt(int integer) { this.integer = integer; } }Then you can sort like this : void sort(LinkedInt linkedInt) { if (linkedInt == null || linkedInt.next == null) return; sort(linkedInt.next); LinkedInt current = linkedInt; int temp; while (current.next != null && current.integer > current.next.integer) { temp = current.integer; current.integer = current.next.integer; current.next.integer = temp; current = current.next; } }It sorts, it is recursive, it have only one parameter and not depends on other method. The idea : we sort the queue first, then since queue is sorted, it is easy to push the value to the good place. Hope it's Help JHelp 
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