Title
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.
题目大意：给一个数组，degree表示为出现次数出现最多的次数，要找的就是出现次数最多元素索引长度的最小值
Solution
//cpp:
class Solution {
public:
int findShortestSubArray(vector<int>& nums) {
unordered_map<int, int> left, right, count;
int degree = 0;
for (int i = 0; i < nums.size(); i++) {
if (!left.count(nums[i])) {
left[nums[i]]=i;
}
right[nums[i]] = i;
count[nums[i]]++;
degree = max(degree, count[nums[i]]);
}
int ans = nums.size();
for (auto x : count) {
if (x.second == degree) {
ans = min(ans, right.at(x.first) - left.at(x.first) + 1);
}
}
return ans;
}
};