【DataStructure】Some useful methods about linkedList(三)

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 should be completely independent of each other. Changing one list should 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: