496. Next Greater Element I LeetCode
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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
andnums2
are unique. - The length of both
nums1
andnums2
would not exceed 1000.
解析:。。。。这还需要么,,,,直接贴答案了
class Solution {public: vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) { vector<int> re; int i=0; for(;i<findNums.size();i++) { int k=0; for(;k<nums.size();k++) if(nums[k]==findNums[i]) break; for(;k<nums.size();k++) { if(findNums[i]<nums[k]) { re.push_back(nums[k]); break; } } if(k==nums.size()) re.push_back(-1); } return re; }};
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