题目描述:
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.
要完成的函数:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums)
说明:
1、给定两个vector,比如[3,2]和[2,4,3,1],第一个vector是第二个的子集。要求输出第二个vector中比第一个vector中某个元素大,并且位置上也在其右边的数。要是没找到的话,输出-1。
比如我们给出的例子中,3的next greater element没有,所以我们输出-1。2的next greater element是4,所以输出4。
2、理解题意之后,一般想法是双重循环。
我们可以设定第一个vector为vector1,第二个为vector2。
对于每个vector1中的元素,遍历vector2,找到其位置,在其位置右边继续找,直到找到比它大的数值。
双重循环,这道题可解。
代码如下: