剑指Offer之Java算法习题精讲链表专题篇

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;
  }
}

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原文链接：https://blog.csdn.net/wai_58934/article/details/123465135

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