Next Greater Element II Problem
Description
LeetCode Problem 503.
Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.
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, return -1 for this number.
Example 1:
1
2
3
4
5
Input: nums = [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.
Example 2:
1
2
Input: nums = [1,2,3,4,3]
Output: [2,3,4,-1,4]
Constraints:
- 1 <= nums.length <= 10^4
- -10^9 <= nums[i] <= 10^9
Sample C++ Code
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
class Solution {
public:
vector<int> nextGreaterElements(vector<int>& nums) {
stack<int> st;
int idx, len = nums.size();
vector<int> ans(len, 0);
for (int i = 0; i < 2*len; i ++) {
idx = i % len;
while (!st.empty()) {
if (nums[st.top()] < nums[idx]) {
ans[st.top()] = nums[idx];
st.pop();
} else {
break;
}
}
if (i < len)
st.push(idx);
}
while (!st.empty()) {
ans[st.top()] = -1;
st.pop();
}
return ans;
}
};