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【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]; } }
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原文地址：http://www.cnblogs.com/lucky-star-star/p/5005808.html
