leetcode day8

【83】 Remove Duplicates from Sorted List

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

思路： 使用一个临时指针next来进行while循环链表，当碰到相同的值时，指针的指针域指向next.next.next向前推进 ,如果碰到不同的，则next = next.next,让next等于当前不同的新节点。使用迭代和递归的时间复杂度是一样的‘

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head==null||head.next==null){
            return head;
        }
        /*//recursive way
        head.next = deleteDuplicates(head.next);
        return head.val == head.next.val ? head.next : head;*/
        
        //iterative way
         ListNode node = head;
        while(node.next!=null){
          if(node.val==node.next.val){
              node.next = node.next.next;
          }else{
              node = node.next;
          } 
         }
        return head;
        
    }
}

【70】Climbing Stairs

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

思路：这个是斐波那契数列问题，只是前两个数是1、2.最好别用斐波那契额递归，时间会超出限制，用数组处理，只需明白思想是，后个数是前两个数的和

public class Solution {

public int climbStairs(int n) {
    if(n == 0 || n == 1 || n == 2){return n;}
    int[] mem = new int[n];
    mem[0] = 1;
    mem[1] = 2;
    for(int i = 2; i < n; i++){
        mem[i] = mem[i-1] + mem[i-2];
    }
    return mem[n-1];
}
}