You are given two arrays (without duplicates) nums1 and nums2 where nums1's elements are a 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.
思路
num1 是 num2 的子集，从 num2 中找到 num1 中指定元素的下一个较大的值并返回，若没有，则返回 -1。循环获取 nums1 中的每个元素，并在 nums2 中查找返回小标迭代器，然后记录下位置按规则向后遍历寻找。
class Solution {
public:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
vector<int> res;
for(auto &elem : findNums) {
auto it=find(nums.begin(),nums.end(),elem);
if(++it==nums.()) {
res.();
} {
(;it!=nums.();++it) {
(*it>elem) {
res.(*it);
;
}
}
(it==nums.()) res.();
}
}
res;
}
};
