LeetCode解题报告—— Reverse Linked List II & Restore IP Addresses & Unique Binary Search Trees II
1. Reverse Linked List II
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given
1->2->3->4->5->NULL, m = 2 and n = 4,
return
1->4->3->2->5->NULL.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
思路:链表移动的题目,要求遍历一遍完成。可以用这个移动模式来解决,对于要移动的区间m~n,往右移动第一个元素,后面的元素添加到最前面,比如2-3-4,左移一次是3-2-4,2再往右移是4-3-2,这样就完成了链表的逆置。
public ListNode reverseBetween(ListNode head, int m, int n) { if(head == null) return null; ListNode dummy = new ListNode(0); // create a dummy node to mark the head of this list dummy.next = head; ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing for(int i = 0; i<m-1; i++) pre = pre.next; ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversed ListNode then = start.next; // a pointer to a node that will be reversed // 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3 // dummy-> 1 -> 2 -> 3 -> 4 -> 5
// 难点在于具体怎么实现那个移动模式 for(int i=0; i<n-m; i++) { start.next = then.next; then.next = pre.next; pre.next = then; then = start.next; } // first reversing : dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4 // second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish) return dummy.next; }
2. Restore IP Addresses
Given a string containing only digits, restore it by returning all possible valid IP address combinations.
For example:
Given
"25525511135",
return
["255.255.11.135", "255.255.111.35"]. (Order does not matter)
思路:首先要明白的是什么是一个valid ip地址,即四位中的每一个少于3位数,值小于255,并且不出现01(应该只有一个1),00这种情形。那么可以用穷举,针对所有可能的情况来遍历判断即可。
public class Solution { public List<String> restoreIpAddresses(String s) { List<String> res = new ArrayList<String>(); int len = s.length(); for(int i = 1; i<4 && i<len-2; i++){ for(int j = i+1; j<i+4 && j<len-1; j++){ for(int k = j+1; k<j+4 && k<len; k++){ String s1 = s.substring(0,i), s2 = s.substring(i,j), s3 = s.substring(j,k), s4 = s.substring(k,len); if(isValid(s1) && isValid(s2) && isValid(s3) && isValid(s4)){ res.add(s1+"."+s2+"."+s3+"."+s4); } } } } return res; } public boolean isValid(String s){ if(s.length()>3 || s.length()==0 || (s.charAt(0)=='0' && s.length()>1) || Integer.parseInt(s)>255) return false; return true; } }
3. Unique Binary Search Trees II
Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1...n.
For example,
Given n = 3, your program should return all 5 unique BST's shown below.
1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3
思路:想到了用递归来解这题,但是具体下手时却遇到了各种问题,现在感觉与树相关的题目,一大部分都可以用递归来解决。
public class Solution { public List<TreeNode> generateTrees(int n) { return genTrees(1,n); }
// 首先要明确的是这个方法返回的是一个集合,即所有可能的BST的根节点,对于它的子递归,返回的是所有可能的左子树和右子树的根节点,比如对于题目中给出的例子
第一层返回的集合中的元素是1,3,3,2,1,第二层的话,对于3来说,返回的集合中的元素应该是3的左子树的根节点2,1和右子树的根节点 null,null public List<TreeNode> genTrees (int start, int end) { List<TreeNode> list = new ArrayList<TreeNode>();
// 递归的返回条件,当给子排序二叉树的start>end时,够造不了,所以将null加入到集合中 if(start>end) { list.add(null); return list; } // 递归的返回条件,一层层递归下来会走到这里。如果相等,则子排序二叉树就是start,也就是叶节点 if(start == end){ list.add(new TreeNode(start)); return list; } // 这里的构造思路是这样的,先去找当前根节点的所有可能的左子树根节点集合和右子树根节点集合,再构造这个当前根节点,再将这个二叉树形串起来 List<TreeNode> left,right; for(int i=start;i<=end;i++) { // 从start到end遍历所有可能的当前根节点,并对每一根节点寻找所有可能的左子树根节点集合和右子树根节点集合 left = genTrees(start, i-1); // 对左子树递归,返回所有可能的根节点集合 right = genTrees(i+1,end); // 对右子树递归,返回所有可能的根节点集合 for(TreeNode lnode: left) // 连接处于递归同一层的左子树根节点和右子树根节点 { for(TreeNode rnode: right) { TreeNode root = new TreeNode(i); //从低向上,一层层将这个排序二叉树构造起来 root.left = lnode; root.right = rnode;
// 对于递归底层返回的是所有可能的左,右子树根节点集合(以构造好,就是左右子树指针都赋好了),对于最上一层返回的就是所有合法的二叉排序树的根节点集合了 list.add(root); } } } return list; } }
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