程式設計師程式碼面試指南 —— 連結串列問題(四)
阿新 • • 發佈:2018-11-28
題目:將單鏈表的每K個節點之間逆序
描述:給定一個單鏈表的頭結點head,實現一個調整單鏈表的函式,使得每K個節點之間逆序,如果最後不夠K個節點一組,則不調整最後幾個節點
例如:
連結串列 1—>2—>3—>4—>5—>6—>7—>8—>null k = 3
調整後:3—>2—>1—>6—>5—>4—>7—>8—>null
想法:逆序可以思考棧stack是否可以解決這個問題,同時迴圈和遞迴也可以實現單鏈表的逆序操作
程式碼:
public class Test { public static void main(String[] args) { // init() Node head = new Node(1); Node node1 = new Node(2); Node node2 = new Node(3); Node node3 = new Node(4); Node node4 = new Node(5); Node node5 = new Node(6); Node node6 = new Node(7); Node node7 = new Node(8); head.next = node1; node1.next = node2; node2.next = node3; node3.next = node4; node4.next = node5; node5.next = node6; node6.next = node7; int k = 3; Node res = reverseKNodes(head, k); while (res != null) { System.out.print(res.value + " "); res = res.next; } } private static Node reverseKNodes(Node head, int k) { if (k < 2) { return head; } Node newHead = head; Node cur = head; Node next = null; Stack<Node> stack = new Stack<>(); Node pre = null; while (cur != null) { stack.push(cur); next = cur.next; if (stack.size() == k) { pre = function(pre, next, stack); newHead = newHead == head ? cur : newHead; } cur = next; } return newHead; } private static Node function(Node left, Node right, Stack<Node> stack) { Node cur = stack.pop(); if (left != null) { left.next = cur; } Node next = null; while (!stack.isEmpty()) { next = stack.pop(); cur.next = next; cur = next; } cur.next = right; return cur; } }
題目:給定一個無序單鏈表的頭節點head,刪除其中值重複出現的節點
描述:例如 1—>2—>3—>3—>4—>4—>2—>1—>1—>null
刪除其中的重複節點之後為1—>2—>3—>4—>null
思路:迴圈遍歷刪除,或者散列表刪除,這裡使用迴圈遍歷刪除,類似於選擇排序
private static Node function(Node head) { Node cur = head; Node next = null; Node pre = null; while (cur != null){ next = cur.next; pre = cur; while (next != null){ if(next.value == cur.value){ pre.next = next.next; }else{ pre = pre.next; } next = pre.next; } cur = cur.next; } return head; } }