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.
/** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */ int* nextGreaterElement(int* findNums, int findNumsSize, int* nums, int numsSize, int* returnSize) { *returnSize=findNumsSize; int *res; res=(int *)calloc(*returnSize,sizeof(int*)); int i,j,k; for(i=0;i<findNumsSize;i++) { for(j=0;j<numsSize;j++) { if(findNums[i]==nums[j]) { for(k=j+1;k<numsSize;k++) { if(nums[k]>nums[j]) { res[i]=nums[k]; break; } } if(k==numsSize) res[i]=-1; } } } return res; }