LeetCode——Next Greater Element I

LeetCode——Next Greater Element I

Question

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.

解题思路

想的就是过滤一遍第二个数组，把每个数右边比它大的第一个数存起来，然后遍历第一个数组，为每个元素找到第一个大的元素。

具体实现

class Solution {

public:

    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {

        map<int, int> dict;

        for (int i = 0; i < nums.size(); i++) {

            int flag = 0;

            int j = i + 1;

            for (; j < nums.size(); j++) {

                if (nums[j] > nums[i]) {

                    flag = 1;

                    break;

                }

            }

            if (flag) {

                dict[nums[i]] = nums[j];

            } else {

                dict[nums[i]] = -1;

            }

        }

        vector<int> res;

        for (int i : findNums) {

            res.push_back(dict[i]);

        }

        return res;

    }

};

相关解答中，用到了栈来遍历第二个数组，这样的时间复杂度会降低到O(n)，而以上这个算法的时间复杂度为O(n^2)。

class Solution {

public:

    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {

        stack<int> s;

        unordered_map<int, int> m;

        for (int n : nums) {

            while (s.size() && s.top() < n) {

                m[s.top()] = n;

                s.pop();

            }

            s.push(n);

        }

        vector<int> ans;

        for (int n : findNums) ans.push_back(m.count(n) ? m[n] : -1);

        return ans;

    }

};