[LeetCode] Next Greater Element I
Posted immjc
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了[LeetCode] Next Greater Element I相关的知识,希望对你有一定的参考价值。
You are given two arrays (without duplicates) nums1
and nums2
where nums1
’s elements are subset of nums2
. Find all the next greater numbers for nums1
‘s elements in the corresponding places of nums2
.
The Next Greater Number of a number x in nums1
is the first greater number to its right in nums2
. If it does not exist, output -1 for this number.
Example 1:
Input: nums1 = [4,1,2], nums2 = [1,3,4,2]. Output: [-1,3,-1] Explanation: For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1. For number 1 in the first array, the next greater number for it in the second array is 3. For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
Example 2:
Input: nums1 = [2,4], nums2 = [1,2,3,4]. Output: [3,-1] Explanation: For number 2 in the first array, the next greater number for it in the second array is 3. For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
Note:
- All elements in
nums1
andnums2
are unique. - The length of both
nums1
andnums2
would not exceed 1000.
首先利用map记录nums中元素及其对应的索引,方便查找。接着在遍历findNums中的元素,根据其在map中相应元素的位置查找下一个较大的数。最后如果找到符合要求的数,跳出循环,将这个数字替换res中对应位置的-1。
class Solution { public: vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) { vector<int> res(findNums.size(), -1); unordered_map<int, int> m; for (int i = 0; i != nums.size(); i++) m[nums[i]] = i; for (int i = 0; i != findNums.size(); i++) { int start = m[findNums[i]]; for (int j = start + 1; j != nums.size(); j++) if (nums[j] > findNums[i]) { res[i] = nums[j]; break; } } return res; } }; // 9 ms
以上是关于[LeetCode] Next Greater Element I的主要内容,如果未能解决你的问题,请参考以下文章
[栈] leetcode 503 Next Greater Element II
[leetcode] Next Greater Element
[leetcode-556-Next Greater Element III]
LeetCode 1019. Next Greater Node In Linked List