Description
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.
分析
- 用hash表存放次數,用另一個hash表存放該數值在數組的起始位置和結束位置。最後把degree的最大的幾個數取出來,取最小長度就行了。
代碼
class Solution {
public:
int findShortestSubArray(vector<int>& nums) {
unordered_map<int,int> m;
unordered_map<int,pair<int,int>> pos;
int degree=0;
for(int i=0;i<nums.size();i++){
m[nums[i]]++;
if(m[nums[i]]==1){
pos[nums[i]]={i,i};
}else{
pos[nums[i]].second=i;
}
degree=max(degree,m[nums[i]]);
}
int res=INT_MAX;
for(auto count:m){
if(degree==count.second){
res=min(res,pos[count.first].second-pos[count.first].first+1);
}
}
return res;
}
};