【LeetCode】1019. Next Greater Node In Linked List
【LeetCode】1019. 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.val such 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.
The given list has length in the range [0, 10000].
题意
这题给出一个数列，对于其中的每个数 i，找出该数列中在数 i 后面的第一个比 i 大的数 j，如果不存在这样的数，则把 j 置为0.
思路
这题可以先设置一个数组 nex，其中，nex[i]表示第 i 个数后面第一个比它大的数的下标，如果不存在这样的数，则 nex[i] = 0。
然后，我们可以从后往前看。对于第 i 个数，如果它是最后一个数，则 nex[i] = 0；否则，看第 i+1 个数是否比它大，如果第 i+1 个数比它大，则 nex[i] = i + 1
；否则，接着看第 nex[i] 个数是否比它大，并依次循环。
代码
class Solution {
public:
vector nextLargerNodes(ListNode* head) {
vectorv, ans;
int i, j, k, m, n;
ListNode *temp = head;
v.clear();
ans.clear();
while(temp != NULL) {
v.push_back(temp->val);
temp = temp->next;
}
vectornex(v.size());
nex[v.size()-1] = 0;
for(i=v.size()-2; i>=0; --i) {
j = i + 1;
if(v[j] > v[i]) {
nex[i] = j;
continue;
}
else {
while(j && v[j] <= v[i]) {
j = nex[j];
}
if(j)
nex[i] = j;
else
nex[i] = 0;
}
}
for(i=0; i