leetcode 697.数组的度

[LeetCode] Degree of an Array 数组的度

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.

题解：
这一题看到两个案例才明确知道要做什么。
数组的度是数组中某个数字出现最多的次数，称为数组的度。
这个题目问的是，求最短的子数组的长度，这个最短子数组是包含出现次数最多的那个数字。
所以我们可以遍历找出出现次数最多的那个数，然后找到这个数的出现的首尾位置，就可以求出最短子数组的长度。
先用一个哈希表建立每个数字与出现次数之间的关系，再用另一个哈希表记录每个数字与其首尾出现的位置。
分两步：

1. 找数组的度
2. 通过数组的度，找到首尾位置。

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        int res = INT_MAX, degree = 0;
        unordered_map<int, int> m;
        unordered_map<int, pair<int, int>> pos; // 首尾位置
        for(int i = 0; i < nums.size(); ++i){
            if(++m[nums[i]] == 1){
                pos[nums[i]] = {i,i};
            }else{
                pos[nums[i]].second = i;
            }
            degree = max(degree, m[nums[i]]);
        }
        for(auto a : m){
            if(degree == a.second){
                res = min(res, pos[a.first].second - pos[a.first].first + 1);
            }
        }
        return res;
    }
};