标签：reverse https 渐变 exit number output init tle ISE
原题链接在这里：https://leetcode.com/problems/next-greater-node-in-linked-list/
题目：
We are given a linked list with head as the first node. Let‘s number the nodes in the list: node_1, node_2, node_3, ... etc.
Each node may have a next larger value: for node_i, next_larger(node_i) is the node_j.valsuch that j > i, node_j.val > node_i.val, and j is the smallest possible choice. If such a j does not exist, the next larger value is 0.
Return an array of integers answer, where answer[i] = next_larger(node_{i+1}).
Note that in the example inputs (not outputs) below, arrays such as [2,1,5] represent the serialization of a linked list with a head node value of 2, second node value of 1, and third node value of 5.
Example 1:
Input: [2,1,5]
Output: [5,5,0]
Example 2:
Input: [2,7,4,3,5]
Output: [7,0,5,5,0]
Example 3:
Input: [1,7,5,1,9,2,5,1]
Output: [7,9,9,9,0,5,0,0]
Note:
1 <= node.val <= 10^9 for each node in the linked list.[0, 10000].题解：
类似Next Greater Element I, Next Greater Element II.
利用stack记录从后向前渐渐变小的栈.
reverse list后就可以从后向前iterate list. 遇到一个点，在stack中一直pop到比自己大的点就是next greater element. 如果stack空了也没找到，就是没有.
记下结果后再把当前点push到stack中.
Time Complexity: O(n).
Space: O(n).
AC Java:
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { val = x; } 7 * } 8 */ 9 class Solution { 10 public int[] nextLargerNodes(ListNode head) { 11 if(head == null){ 12 return new int[0]; 13 } 14 15 if(head.next == null){ 16 return new int[]{0}; 17 } 18 19 int len = getLength(head); 20 int [] res = new int[len]; 21 head = reverse(head); 22 Stack<Integer> stk = new Stack<Integer>(); 23 24 for(int i = 0; i<len; i++){ 25 while(!stk.isEmpty() && head.val>=stk.peek()){ 26 stk.pop(); 27 } 28 29 res[len-i-1] = stk.isEmpty() ? 0 : stk.peek(); 30 stk.push(head.val); 31 head = head.next; 32 } 33 34 return res; 35 } 36 37 private ListNode reverse(ListNode head){ 38 if(head == null || head.next == null){ 39 return head; 40 } 41 42 ListNode tail = head; 43 ListNode cur = head; 44 ListNode pre; 45 ListNode temp; 46 while(tail.next != null){ 47 pre = cur; 48 cur = tail.next; 49 temp = cur.next; 50 cur.next = pre; 51 tail.next = temp; 52 } 53 54 return cur; 55 } 56 57 private int getLength(ListNode head){ 58 int res = 0; 59 while(head != null){ 60 head = head.next; 61 res++; 62 } 63 64 return res; 65 } 66 }
LeetCode 1019. Next Greater Node In Linked List
标签：reverse https 渐变 exit number output init tle ISE
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/10967068.html
