496.Next Greater Element I

Next Greater Element I

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.

代码

思路

首先将nums里面的所有的数据放入字典中，然后遍历findNums里面的数据，根据数据在字典里面的位置，往后搜索比其大的元素。

vector<int> nextGreaterElement(vector<int> &findNums, vector<int> &nums) {
unordered_map<int, int> map{};
for (auto iter = nums.begin(); iter != nums.end(); iter++)
map[*iter] = (iter - nums.begin());
for (auto iterFind = findNums.begin(); iterFind != findNums.end(); iterFind++) {
int distance = iterFind - findNums.begin();
int temp = -1;
for (auto iterNums = nums.begin() + map[*iterFind]; iterNums != nums.end(); iterNums++) {
if (*iterNums > *iterFind) {
temp = *iterNums;
break;
}
}
findNums[distance] = temp;
}
for(auto iter=findNums.begin();iter!=findNums.end();iter++)
{
cout<<*iter<<ends;
}
return findNums;
}