[Datastructure] Some useful methods about into list (3)

Method 4: gets the value of element number I

For example, if list is {22, 33, 44, 55, 66, 77, 88, 99}, then get (list, 2) will return 44.

Solution 1:static int get(Node list, int i) {if (i < 0) {throw new IllegalArgumentException();}for (int j = 0; j < i; j++) {if (list == null) {throw new IllegalStateException();}list = list.next;}return list.data;}Solution 2:static int get(Node list, int i){Node p = list;int j = 0;while (j < i && p != null) {++j;p = p.next;}if(p == null){throw new java.util.NoSuchElementException();}return p.data;}

The output is as follows:

Method 5: inserts X as element number I;

For example, if list is {22, 33, 44, 55, 66, 77, 88, 99}, then put (list, 3, 50) will change list to {22, 33, 44, 50, 55, 66, 44, 88, 99 }.

Hint:If I = 0, replace the value of the first node with X, and insert a new node immediately after it that contains the previous fist value.

Solution 1:static void put(Node list, int i, int x) {if (list == null) {throw new java.util.NoSuchElementException("List is Empty");} else if (i ==0) {list.next = new Node(list.data,list);list.data = x;} else {Node p = list;int j = 1;while (j < i && p != null) {++j;p = p.next;}if (p == null){String error = String.format("the list has only %d elements", j-1);;throw new java.util.NoSuchElementException(error);}p.next = new Node(x, p.next);}}

The output is as follows:

Method 6: swap the I element with the J Element

For example, if list is {22, 33, 44, 55, 66, 77, 88, 99}, then swap (list, 2, 5) will change list to {22, 33, 77, 55, 66, 44, 88, 99 }.

static void swap(Node list, int i, int j) {if (i < 0 || j < 0) {throw new IllegalArgumentException();} else if (i == j) {return;}Node p = list, q = list;for (int ii = 0; ii < i; ii++) {if (p == null) {throw new IllegalStateException();}p = p.next;}for (int jj = 0; jj < j; jj++) {if (q == null) {throw new IllegalStateException();}q = q.next;}int pdata = p.data, qdata = q.data;p.data = qdata;q.data = pdata;return;}

The output is as follows:

Method 7: gets a new list that contains all the elements of list1 and list2 in ascending order. list1 and list2 are both in ascending order.

For example, if list1is {22, 33, 55, 88} And list2is {44, 66, 77, 99}, then merged (list1, list2) will return the new list {22, 33, 44, 55, 66, 77, 88, 99 }.

Note that the three lists shocould be completely independent of each other. Changing one list shoshould have no effect upon the others.

static Node merged(Node list1, Node list2) {Node list = new Node(0);Node p = list, p1 = list1, p2 = list2;while (p1 != null && p2 != null) {if (p1.data < p2.data) {p = p.next = new Node(p1.data);p1 = p1.next;} else {p = p.next = new Node(p2.data);p2 = p2.next;}}while (p1 != null) {p = p.next = new Node(p1.data);p1 = p1.next;}while (p2 != null) {p = p.next = new Node(p2.data);p2 = p2.next;}return list.next;}

The output is as follows: