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.
1 class Solution { 2 public: 3 int findShortestSubArray(vector<int>& nums) { 4 unordered_map<int, pair<int, int>> m1; 5 unordered_map<int, int> m2; 6 int count = 0,res = nums.size(); 7 for (int i = 0; i < nums.size(); i++) 8 { 9 m2[nums[i]]++; 10 if (m1.find(nums[i]) == m1.end()) m1[nums[i]] = pair<int, int>(i, i); 11 else m1[nums[i]].second = i; 12 if (m2[nums[i]] > count) count = m2[nums[i]]; 13 } 14 for (auto num : m2) 15 { 16 if (count == num.second) 17 { 18 res = min(res, m1[num.first].second - m1[num.first].first + 1); 19 } 20 } 21 return res; 22 } 23 };
时间: 2024-10-09 18:16:58