LeetCode 503. Next Greater Element II

原题链接在这里：https://leetcode.com/problems/next-greater-element-ii/

题目：

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.

题解：

是Next Greater Element I进阶版.

一般这类带环的一种常见做法是加长一倍. 这里就是把nums 走两边.

 没用HashMap<Integer, Integer>是因为nums里会有duplicate. 这里的Stack<Integer> stk 存的是element index 而不是element, 需要index来填res.

Time Complexity: O(n), n = nums.length.

Space: O(n), res size.

AC Java:

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        if(nums == null || nums.length == 0){
            return nums;
        }

        Stack<Integer> stk = new Stack<>();
        int n = nums.length;
        int [] res = new int[n];
        for(int i = 2 * n - 1; i >= 0; i--){
            int ind = i % n;
            while(!stk.isEmpty() && nums[stk.peek()] <= nums[ind]){
                stk.pop();
            }

            if(i < n){
                res[ind] = stk.isEmpty() ? -1 : nums[stk.peek()];
            }

            stk.push(ind);
        }

        return res;
    }
}

We could also iterate from left to right and have monotonic increasing stack. When encounter current element x is bigger, update all the popped index element next greater as x.

Then push ind = i % n in the stack. 

Time Complexity: O(n).

Space: O(n).

AC Java:

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        if(nums == null || nums.length == 0){
            return nums;
        }

        int n = nums.length;
        int [] res = new int[n];
        Arrays.fill(res, -1);
        Stack<Integer> stk = new Stack<>();
        for(int i = 0; i < 2 * n; i++){
            int ind = i % n;
            while(!stk.isEmpty() && nums[stk.peek()] < nums[ind]){
                res[stk.pop()] = nums[ind];
            }

            stk.push(ind);
        }

        return res;
    }
}

AC C++:

class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        int n = nums.size();
        vector<int> res(n, -1);
        stack<int> stk;

        for(int i = 0; i < n * 2; i++){
            int ind = i % n;
            while(!stk.empty() && nums[stk.top()] < nums[ind]){
                int top = stk.top();
                stk.pop();
                res[top] = nums[ind];
            }

            stk.push(ind);
        }

        return res;
    }
};

AC Python:

class Solution:
    def nextGreaterElements(self, nums: List[int]) -> List[int]:
        n = len(nums)
        stk = []
        res = [-1] * n
        for i in list(range(n)) * 2:
            while stk and nums[stk[-1]] < nums[i]:
                res[stk.pop()] = nums[i]
            stk.append(i)
        return res

跟上Next Greater Element III.