Problem:
Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.
Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.
Example 1:
Input: [1, 2, 2, 3, 1]
Output: 2
Explanation:
The input array has a degree of 2 because both elements 1 and 2 appear twice.
Of the subarrays that have the same degree:
[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
The shortest length is 2. So return 2.
Example 2:
Input: [1,2,2,3,1,4,2]
Output: 6
Note:
- nums.length will be between 1 and 50,000.
- nums[i] will be an integer between 0 and 49,999.
思路：
Solution (C++):
int findShortestSubArray(vector<int>& nums) {
if (nums.empty()) return 0;
int n = nums.size(), res = 0, degree = 0;
unordered_map<int, int> first, count;
for (int i = 0; i < n; ++i) {
if (first.count(nums[i]) == 0) first[nums[i]] = i;
if (++count[nums[i]] > degree) {
degree = count[nums[i]];
res = i - first[nums[i]] + 1;
} else if (count[nums[i]] == degree) {
res = min(res, i - first[nums[i]] + 1);
}
}
return res;
}
性能：
Runtime: 60 ms Memory Usage: 10.9 MB
思路：
Solution (C++):
性能：
Runtime: ms Memory Usage: MB
