链表

BM1:反转链表

给定一个单链表的头结点Head(该头节点是有值的)，长度为n，反转该链表后，返回新链表的表头。
BM1 题目链接
case1:实际如下所示输入输出
输入：{1,2,3}
输出：{3,2,1}

  1. /*
  2. public class ListNode {
  3.     int val;
  4.     ListNode next = null;
  6.     ListNode(int val) {
  7.         this.val = val;
  8.     }
  9. }*/
  10. import java.io.*;
  11. public class Solution {
  12.     public static void ReverseList(ListNode head) throws IOException{
  13.         BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
  14.         String readline = bufferedReader.readLine();
  15.         readline = readline.substring(1,readline.length()-1);
  16.         StringBuffer stringBuffer = new StringBuffer();
  17.         String[] splits = readline.split(",");
  18.         stringBuffer.append("{");
  19.         for(int i = splits.length-1 ; i >= 0; i--){
  20.             if(i == 0){
  21.                 stringBuffer.append(splits[i]);
  22.             }else{
  23.                 stringBuffer.append(splits[i]+",");
  24.             }
  25.         }
  26.         stringBuffer.append("}");
  27.         System.out.println(stringBuffer.toString());
  28.     }
  29. }

case2:链表反转

  1. /*
  2. public class ListNode {
  3.     int val;
  4.     ListNode next = null;
  6.     ListNode(int val) {
  7.         this.val = val;
  8.     }
  9. }*/
  10. public class Solution {
  11.     public ListNode ReverseList(ListNode head) {
  12.         if(head == null)  return null;
  13.         ListNode pre = null;
  14.         ListNode cur = head;
  15.         while(cur != null){
  16.             ListNode tmp = cur.next;
  17.             cur.next = pre;
  18.             pre = cur;
  19.             cur = tmp;
  20.         }
  21.         return pre;
  22.     }
  23. }

二分查找/排序

BM17:二分查找-1

实现无重复数字的升序数组的二分查找
BM17 题目链接

  1. import java.util.*;
  3. public class Solution {
  4.     /**
  5.      * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
  6.      *
  7.      * 
  8.      * @param nums int整型一维数组 
  9.      * @param target int整型 
  10.      * @return int整型
  11.      */
  12.     public int search (int[] nums, int target) {
  13.         // write code here
  14.         int low = 0, high = nums.length-1;
  15.         while(low <= high){
  16.             int mid = low + (high - low) / 2;
  17.             if(nums[mid] == target){
  18.                 return mid;
  19.             }else if(nums[mid] > target){
  20.                 high = mid - 1;
  21.             }else{
  22.                 low = mid+1;
  23.             }
  24.         }
  25.         return -1;
  26.     }
  27. }

二叉树

BM23:二叉树的前序遍历

给你二叉树的根节点 root ，返回它节点值的 前序 遍历。
BM23 题目链接