697 Degree of an Array 数组的度
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:
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.
题目描述:
给定一个非空且只包含非负数的整数数组 nums, 数组的度的定义是指数组里任一元素出现频数的最大值。
你的任务是找到与 nums 拥有相同大小的度的最短连续子数组，返回其长度。
示例 :
示例 1:
输入: [1, 2, 2, 3, 1]
输出: 2
解释:
输入数组的度是2，因为元素1和2的出现频数最大，均为2.
连续子数组里面拥有相同度的有如下所示:
[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
最短连续子数组[2, 2]的长度为2，所以返回2.
示例 2:
输入: [1,2,2,3,1,4,2]
输出: 6
注意:
nums.length 在1到50,000区间范围内。
nums[i] 是一个在0到49,999范围内的整数。
思路:
用一个 map记录下各个元素出现的位置, 然后找到出现次数最多的元素, 即为数组的度, 然后找到数组的度对应的元素的起始位置即可
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
class Solution
{
public:
int findShortestSubArray(vector& nums)
{
unordered_map> m;
for (int i = 0; i < nums.size(); i++) m[nums[i]].push_back(i);
int degree = 1, result = INT_MAX;
for (auto i : m) degree = max(degree, int(i.second.size()));
for (auto i : m) if (degree == i.second.size()) result = min(result, i.second.back() - i.second[0] + 1);
return result;
}
};
Java:
class Solution {
public int findShortestSubArray(int[] nums) {
Map> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
if (map.get(nums[i]) == null) {
List list = new ArrayList<>();
list.add(i);
map.put(nums[i], list);
} else map.get(nums[i]).add(i);
}
int degree = 1, result = Integer.MAX_VALUE;
for (Integer key : map.keySet()) degree = Math.max(degree, map.get(key).size());
for (Integer key : map.keySet()) if (degree == map.get(key).size()) result = Math.min(result, map.get(key).get(map.get(key).size() - 1) - map.get(key).get(0) + 1);
return result;
}
}
Python:
class Solution:
def findShortestSubArray(self, nums: List[int]) -> int:
c = collections.Counter(nums)
d = max(c.values())
arr = [k for k, v in c.items() if v == d]
return min([len(nums) - nums[::-1].index(n) - nums.index(n) for n in arr])