5.15 - Stacks and Queues
Posted AugusKong
tags:
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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 <= 0) {
//if the input is null or the head points to the middle of the list
return new Result(head, true);
} else if (length == 1) {
// if the middle + 1 of the list
return new Result(head.next, true);
}
// 递归的调用
Result res = isPalindromeRecurse(head.next, length - 2);
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
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