503 Next Greater Element II
Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, output -1 for this number.
Example 1:
Input:
[1,2,1]
Output:
[2,-1,2]
Explanation:
The first 1's next greater number is 2;
The number 2 can't find next greater number;
The second 1's next greater number needs to search circularly, which is also 2.
Note:The length of given array won't exceed 10000.
The Idea: I hashed the locations for the second list, and use that as the starting point of traversing the first list until a greater element is reached.
vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2)
{
vector<int> next_max_ar;
next_max_ar.reserve(nums1.size());
unordered_map<int, int> num_loc;
num_loc.reserve(nums2.size());
for (int i = 0; i < nums2.size(); i++)
num_loc.insert({ nums2[i], i });
for (int i = 0; i < nums1.size(); i++) {
int cur_ele = nums1[i];
int start_loc = num_loc[cur_ele] + 1;
int next_max = -1;
for (int j = start_loc; j < nums2.size(); j++) {
if (nums2[j] > cur_ele) {
next_max = nums2[j];
break;
}
}
next_max_ar.push_back(next_max);
}
return next_max_ar;
}
Last modified 3yr ago