[每日 LeetCode] 496. Next Greater Element I

原文链接 [每日 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.

本题题意是给出两个数组，其中有一个是另一个的子数组，找出子数组中的每个元素在原数组中比这个数大的第一个数。

思路一：暴力解法。依次遍历子数组中的元素，找到每个元素在原数组中的位置，然后在原数组中改数的后面查找比此数大的数，有则返回第一个大的数，没有则返回-1。

思路二：使用 unordered_map 和 stack 结构，首先遍历原数组，将每个元素及此元素后面第一个大的数组成键值对，然后遍历子数组，在 map 中查找键值，若存在则返回 value 值，不存在则返回-1。

C++ 代码）（思路一）

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        vector<int> res(nums1.size());
        for (int i = 0; i < nums1.size(); ++i) {
            int j = 0, k = 0;
            for (; j < nums2.size(); ++j) {
                if (nums2[j] == nums1[i]) 
                    break;
            }
            for (k = j + 1; k < nums2.size(); ++k) {
                if (nums2[k] > nums2[j]) {
                    res[i] = nums2[k];
                    break;
                }
            }
            if (k == nums2.size()) 
                res[i] = -1;
        }
        return res;
    }
};

运行时间：12ms

运行内存：8.6M

C++ 代码（思路二）

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        stack<int> s;
        unordered_map<int, int> m;
        for (int n : nums2) {
            while (s.size() && s.top() < n) {
                m[s.top()] = n;
                s.pop();
            }
            s.push(n);
        }
        vector<int> ans;
        for (int n : nums1) 
	    ans.push_back(m.count(n) ? m[n] : -1);
        return ans;
    }
};

运行时间：12ms

运行内存：9.7M