[Leetcode Daily Question 2021/10/26] 496. The next bigger element I [Simple] Advanced: Monotone Stack + Hash Table

The question comes from leetcode, and the solution and ideas represent only my personal opinions. Portal.
Difficulty: Easy
Time: 10min

topic

Give you two arrays nums1 and nums2 without duplicate elements, where nums1 is < a subset of i=4>. nums2

Please find the next greater value of each element in nums1 in nums2.

nums1xThe next larger element of the number in at the corresponding position. . If it does not exist, output An element larger than x in is the element to the right of the corresponding position in nums2x-1

Example 1:

输入: nums1 = [4,1,2], nums2 = [1,3,4,2].
输出: [-1,3,-1]
解释:
    对于 num1 中的数字 4 ，你无法在第二个数组中找到下一个更大的数字，因此输出 -1 。
    对于 num1 中的数字 1 ，第二个数组中数字1右边的下一个较大数字是 3 。
    对于 num1 中的数字 2 ，第二个数组中没有下一个更大的数字，因此输出 -1 。

Example 2:

输入: nums1 = [2,4], nums2 = [1,2,3,4].
输出: [3,-1]
解释:
    对于 num1 中的数字 2 ，第二个数组中的下一个较大数字是 3 。
    对于 num1 中的数字 4 ，第二个数组中没有下一个更大的数字，因此输出 -1 。

Hint: All the integers in
1 <= nums1.length <= nums2.length <= 1000
0 <= nums1[i], nums2[i] <=10^4
nums1 and nums2 are different from each other All integers in also appear in
nums1nums2

Advanced: Can you design a solution with time complexity O(nums1.length + nums2.length)?

Ideas

1. Violent solution. I won’t write this anymore.
2. Monotone stack + hash table.

Monotone stack + hash table

In fact, this is a type of question - monotonic stack. So record it.

Identification keywords:
Next, bigger, smaller

---

Because,
nums1 is a subset of nums2.
nums1 and nums are both unordered.
The result is the next larger element of all elements in nums1.

Then,
we need to use a map recordnums1 element of The next larger element.
Moreover, using a monotonic stack, we can find the next greater elementnums2 of all elements in .

---

Monotonic stack operation process (finding the next larger number):
For all elements of nums2:

1. If the stack is empty, push it当前元素.
2. If the stack is not empty, 栈顶元素 is compared with 当前元素.

• Result栈顶元素 >= 当前元素
  Entering当前元素.
• If栈顶元素 < 当前元素
  (当前元素 is, 栈顶元素 a>The next larger element)
  Execute the loop and pop all elements smaller than 当前元素 from the stack element (itsnext larger element is also当前元素).

code

class Solution {
    
    
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
    
    
        int left = 0;
        int rihgt = 0;
        int n1 = nums1.size();
        int n2 = nums2.size();
        vector<int> ans(n1,-1);

        unordered_map<int,int> mmap;
        stack<int> s;
        for(int i=0;i<n2;i++){
    
    
            if(s.empty()){
    
    
                s.push(nums2[i]);
            }else if(s.top() < nums2[i]){
    
    
                while(!s.empty() && s.top() < nums2[i]){
    
    
                    int num = s.top();
                    s.pop();
                    mmap[num] = nums2[i];
                }
                s.push(nums2[i]);
            }else if(s.top() >= nums2[i]){
    
    
                s.push(nums2[i]);
            }
        }
        for(int i=0;i<n1;i++){
    
    
            if(mmap.count(nums1[i]) == 0){
    
    
                continue;
            }
            ans[i] = mmap[nums1[i]];
        }
        return ans;
    }
};

---

Optimize structural code

class Solution {
    
    
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
    
    
        int left = 0;
        int rihgt = 0;
        int n1 = nums1.size();
        int n2 = nums2.size();
        vector<int> ans(n1,-1);

        unordered_map<int,int> mmap;
        stack<int> s;
        for(int i=0;i<n2;i++){
    
    
            while(!s.empty() && s.top() < nums2[i]){
    
    
                int num = s.top();
                s.pop();
                mmap[num] = nums2[i];
            }
            s.push(nums2[i]);
        }
        for(int i=0;i<n1;i++){
    
    
            if(mmap.count(nums1[i]) == 0){
    
    
                continue;
            }
            ans[i] = mmap[nums1[i]];
        }
        return ans;
    }
};

Algorithmic complexity

Time efficiency： $O(n_1+n_2)$, inside$n_1$为nums1's length, $n_2$为nums2's length. $O\left(n_1\right)$Using the generated answer,$O\left(n_2\right)$usesearch for the single largest element.
Space power： $O(n_1+n_2)$, inside$n_1$为nums1's length, $n_2$为nums2's length. $O\left(n_1\right)$用于哈子表，$O\left(n_2\right)$Using the training station.