剑指Offer之Java算法习题精讲链表专题篇
作者：明天一定.
跟着思路走，之后从简单题入手，反复去看，做过之后可能会忘记，之后再做一次，记不住就反复做，反复寻求思路和规律，慢慢积累就会发现质的变化
题目一
解法
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public int getDecimalValue(ListNode head) {
int[] arr = new int[31];
int index = 0;
int ans = 0;
while(head!=null){
arr[index] = head.val;
index++;
head = head.next;
}
for(int i = 0;i<index;i++){
if(arr[i]==1){
ans+=(1<<(index-1-i));
}
}
return ans;
}
}
题目二
解法
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public int[] reversePrint(ListNode head) {
int index = 0;
ListNode h = head;
while(head!=null){
head = head.next;
index++;
}
int[] arr = new int[index];
while(h!=null){
arr[index-1] = h.val;
index--;
h = h.next;
}
return arr;
}
}
题目三
解法
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode node = new ListNode(-1);
ListNode ans = node;
while(l1!=null&&l2!=null){
if(l1.val<=l2.val){
node.next = l1;
l1 = l1.next;
}else{
node.next = l2;
l2 = l2.next;
}
node = node.next;
}
if(l1!=null){
node.next = l1;
}
if(l2!=null){
node.next = l2;
}
return ans.next;
}
}
到此这篇关于剑指Offer之Java算法习题精讲链表专题篇的文章就介绍到这了,更多相关Java 链表内容请搜索脚本之家以前的文章或继续浏览下面的相关文章希望大家以后多多支持脚本之家！