5.15 - Stacks and Queues

Solution1: Reverse and Compare

翻转整个Linked List, 然后用两个指针进行逐一比对

Solution2: Iterative Approach -> use Stack

boolean isPalindrome(ListNode head) {
    ListNode fast = head;
    ListNode slow = head;
    
    Stack<Integer> stack = new Stack<Integer>();

    while (fast != null && fast.next != null) {
        stack.push(slow.data);
        slow = slow.next;
        fast = fast.next.next;
    }

// Has odd number of elements, so skip the middle element 
    if (fast != null) {
        slow = slow.next;
    }

    while (slow != null) {
        int top = stack.pop().intValue();

        /* If values are different, then it's not a palindrome */
        if (top != slow.data) {
            return false;
        }
        slow = slow.next;
    }
    return true;
}

```
Recursive Approach
Q: How recursive function would work?
A:

Q: How can we get to the end of the list? How can we compare node i with node n - i?

从head 开始遍历，但是从中间开始做compare

class Result {
    public ListNode node;
    public boolean result;
}

/* Result当中 {
    node 存储的是当前遍历的指针位置    
    result表示当前比较的结果 
}

*/
boolean isPalindrome(ListNode head) {
    int length = getLength(head);
    Result p = isPalindromeRecurse(head, length);
    return p.result;
}

Result isPalindromeRecurse(ListNode head, int length) {
    
    // 递归的出口 
    if (head == null || length <= ) { 
        //if the input is null  or the head points to the middle of the list
        return new Result(head, true);
    } else if (length == ) { 
        // if the middle + 1 of the list  
        return new Result(head.next, true);
    }
    
    // 递归的调用
    Result res = isPalindromeRecurse(head.next, length - );
    if (!res.result || res.node == null) {
        return res;
    }

    // result
    res.result = (head.data == res.node.data);
    /* Return corresponding node. */
    //在递归出口之前 将指针右移
    res.node = res.node.next;
    
    return res;
}

160. Intersection of Two Linked Lists (cc189 2.7)

Q: insersection 节点之后的ListNode 全部重合，
至少要遍历一遍linkedlist
只给了两个头结点，主要问题在于当两个list的长度不一致的时候，如何进行intersection的确定？ 链表的遍历是否要同步？ 这样会出现headA超前headB的情况