leetcode 496. Next Greater Element I

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.

My solution

• 最笨的方式显然是完全暴力搜索, 仔细考察浪费计算量的地方.
• 像discuss code中那样利用stack!!!!

思路: [1,3,4,2,5] 而言, 4和2的next greater element都是5. 4 要往后找, 如果下下个是, 那么这个数也是2的, 所以就转化为用stack把4存下来, 直接找2, 找完2的,再来和stack的元素对比, 找到的元素都出栈!!!
这么看来, 对于需要保存历史, 并且是有序信息的(先问题转为后问题的再比较), 可以巧妙的利用stack

//class Solution {
//public:
//    vector nextGreaterElement(vector &findNums, vector &nums) {
//        vector res(findNums.size(), -1);
//        for (int i = 0; i < findNums.size(); ++i) {
//            for (int j = 0; j < nums.size(); ++j) {
//                if (findNums[i] == nums[j]) {
//                    for (int k = j + 1; k < nums.size(); ++k) {
//                        if (nums[k] > nums[j]) {
//                            res[i] = nums[k];
//                            break;
//                        }
//                    }
//                }
//            }
//        }
//        return res;
//    }
//};
class Solution {
public:
    vector nextGreaterElement(vector &findNums, vector &nums) {
        vector res(findNums.size(), -1);
        stack s;
        unordered_map mp;
        for (int i = 0; i < nums.size(); ++i) {
            while (!s.empty() && s.top() < nums[i]) {
                mp[s.top()] = nums[i];
                s.pop();
            }
            s.push(nums[i]);
        }
        for (int i = 0; i < findNums.size(); ++i) {
            if (mp[findNums[i]] != 0) res[i] = mp[findNums[i]];
        }
        return res;
    }
};

Discuss

• C++ stack + unordered_map

class Solution {
public:
    vector nextGreaterElement(vector& findNums, vector& nums) {
        stack s;
        unordered_map m;
        for (int n : nums) {
            while (s.size() && s.top() < n) {
                m[s.top()] = n;
                s.pop();
            }
            s.push(n);
        }
        vector ans;
        for (int n : findNums) ans.push_back(m.count(n) ? m[n] : -1);
        return ans;
    }
};

Reference

-leetcode 496. Next Greater Element I