android-notes/blogs/Algorithm/剑指 Offer/链表相关.md

链表相关

06. 从头到尾打印链表

class Solution {
    public int[s] reversePrint(ListNode head) {
        ListNode temp = head;
        int size = 0;
        while (temp != null) {
            temp = temp.next;
            size++;
        }
        int[] result = new int[size];
        int index = size - 1;
        while (head != null) {
            result[index--] = head.val;
            head = head.next;
        }
        return result;
    }
}

class Solution {
    public int[] reversePrint(ListNode head) {
        Stack<ListNode> stack = new Stack<ListNode>();
        ListNode temp = head;
        while (temp != null) {
            stack.push(temp);
            temp = temp.next;
        }
        int size = stack.size();
        int[] print = new int[size];
        for (int i = 0; i < size; i++) {
            print[i] = stack.pop().val;
        }
        return print;
    }
}

22. 链表中倒数第 k 个节点

class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        int length = 0;
        ListNode temp = head;
        while (temp != null) {
            temp = temp.next;
            length++;
        }
        for (int i = 0; i < length - k; i++) {
            head = head.next;
        }
        return head;
    }
}

class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        ListNode h0 = head;
        for (int i = 0; i < k; i++) {
            h0 = h0.next;
        }
        while (h0 != null) {
            h0 = h0.next;
            head = head.next;
        }
        return head;
    }
}

24. 反转链表

    class Solution {
        public ListNode reverseList(ListNode head) {
            if (head == null || head.next == null) {
                return head;
            }
            ListNode h1 = head;
            ListNode h2 = head.next;
            ListNode h3 = null;
            h1.next = null;
            while (h2 != null) {
                h3 = h2.next;
                h2.next = h1;
                h1 = h2;
                h2 = h3;
            }
            return h1;
        }
    }

    class Solution {
        public ListNode reverseList(ListNode head) {
            // 递归终止条件是当前为空，或者下一个节点为空
            if (head == null || head.next == null) {
                return head;
            }
            // 这里的 h1 就是最后一个节点
            ListNode h1 = reverseList(head.next);
            // 如果链表是 1->2->3->4->5，那么此时的 cur 就是 5
            // 而 head 是4，head的 下一个是 5，下下一个是空
            // 所以 head.next.next 就是 5->4
            head.next.next = head;
            // 防止链表循环，需要将 head.next 设置为空
            head.next = null;
            // 每层递归函数都返回 h1，也就是最后一个节点
            return h1;
        }
    }