496. Next Greater Element I

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.

方法1: 单调栈

思路：

首先建立一下nums1中数字和index的映射，然后遍历nums2建立单调递减栈，这里储存的可以是nums2也可以是num本身。根据这个单调栈的性质，每一次遇到比栈顶更大的数字，开始pop，并且在nums1中查找这些数字，如果存在于nums1，通过hash找到坐标，将相应位置替换为nums2[i]。如果比栈顶小，说明栈内元素尚未遇到next greater，继续排队等待。新元素在排除掉之前比它小的所有栈顶元素后（while loop）入栈。

Complexity

Time complexity: O(n), n is length of nums2。
Space complexity: O(n)。

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        vector<int> res(nums1.size(), -1);
        unordered_map<int, int> hash;
        stack<int> st;
        for (int i = 0; i < nums1.size(); i++) hash[nums1[i]] = i;       
        for (int i = 0; i < nums2.size(); i++) {
            while (!st.empty() && nums2[st.top()] < nums2[i]) {
                int top = st.top(); st.pop();
                if  (hash.find(nums2[top]) == hash.end()) continue;
                res[hash[nums2[top]]] = nums2[i];
            }
            st.push(i);
        }
        return res;
    }
};

另一种写法，hashmap 用来保存num和next greater的映射，栈保存num2中的数字本身。

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        vector<int> res;
        stack<int> st;
        unordered_map<int, int> m;
        for (int num : nums) {
            while (!st.empty() && st.top() < num) {
                m[st.top()] = num; st.pop();
            }
            st.push(num);
        }
        for (int num : findNums) {
            res.push_back(m.count(num) ? m[num] : -1);
        }        
        return res;
    }
};