标签:rip empty return logs push stack ret ems iuc
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:
nums1
and nums2
are unique.nums1
and nums2
would not exceed 1000.Solution:
class Solution { public: vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) { vector<int> result; stack<int> s; map<int, int> m; for (int i = nums.size() - 1; i >= 0; i--) { while (!s.empty() && s.top() <= nums[i]) s.pop(); m[nums[i]] = s.empty() ? -1 : s.top(); s.push(nums[i]); } for (int i = 0; i < findNums.size(); i++) result.push_back(m[findNums[i]]); return result; } };
题目直达:https://leetcode.com/problems/next-greater-element-i/#/description
答案直达:http://www.liuchuo.net/archives/3197
标签:rip empty return logs push stack ret ems iuc
原文地址:http://www.cnblogs.com/SapphireCastle/p/6759625.html