496. Next Greater Element I*

496. Next Greater Element I*

https://leetcode.com/problems/next-greater-element-i/

Title Description

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.

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.

Note:

• All elements in nums1 and nums2 are unique.
• The length of both nums1 and nums2 would not exceed 1000.

C ++ implementation 1

Find num1each number in the nums2next greater number in. First a hash table is stored nums2in the index assigned to each number, and then look for nums1the number in each nums2location idx, and finally find the Next Greater Number.

This question can also use the stack to do, see C++ 实现 2, stack usage is more common, for example, can resolve 1019. Next Greater Node In Linked List ** this question.

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int, int> record;
        for (int i = 0; i < nums2.size(); ++ i) record[nums2[i]] = i;
        vector<int> res;
        for (auto &n : nums1) {
            auto idx = record[n];
            auto greater = n;
            for (int i = idx + 1; i < nums2.size(); ++ i) {
                if (nums2[i] > greater) {
                    greater = nums2[i];
                    break;
                }
            }
            if (greater != n) res.push_back(greater);
            else res.push_back(-1);
        }
        return res;
    }
};

2 in C ++

Use stack to achieve. Hash table recordrecords nums2for each element in the presence of the NEG, their corresponding positions NEG If the traversal nums2process, temporarily NEG not encountered, the elements should be added to the stack.

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int, int> record;
        vector<int> res(nums1.size(), -1);
        stack<int> st;
        // record 保存了 nums2 中每个存在 NGE 的元素对应的 NGE 的索引
        for (int i = 0; i < nums2.size(); ++ i) {
            while (!st.empty() && nums2[i] > nums2[st.top()]) {
                record[nums2[st.top()]] = i;
                st.pop();
            }
            st.push(i);
        }
        for (int i = 0; i < nums1.size(); ++ i)
            if (record.count(nums1[i]))
                res[i] = nums2[record[nums1[i]]]; // 注意 record 中保存的是索引
        return res;
    }
};