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496. 求下一个更大的元素 Next Greater Element I
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.
public class Solution {
public int[] NextGreaterElement(int[] findNums, int[] nums) {
int[] resultArr = new int[findNums.Length];
Dictionary<int, int> indexDict = new Dictionary<int, int>();
for (int i = 0; i < nums.Length;i++ )
{
indexDict[nums[i]] = i;
}
for (int i = 0; i < findNums.Length; i++)
{
int index = indexDict[findNums[i]];
int num = -1;
for (; index < nums.Length; index++)
{
if (nums[index] > findNums[i])
{
num = nums[index];
break;
}
}
resultArr[i] = num;
}
return resultArr;
}
}
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496. 求下一个更大的元素 Next Greater Element I
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