[LeetCode] 961. N-Repeated Element in Size 2N Array

Description

In a array A of size 2N, there are N+1 unique elements, and exactly one of these elements is repeated N times.

Return the element repeated N times.

Example 1:

Input: [1,2,3,3]
Output: 3

Example 2:

Input: [2,1,2,5,3,2]
Output: 2

Example 3:

Input: [5,1,5,2,5,3,5,4]
Output: 5

Note:

1. 4 <= A.length <= 10000
2. 0 <= A[i] < 10000
3. A.length is even

---

Analyse

数组A一共2N个数，其中一个数出现了N次，整个数组A中共有N+1个不一样的数

很容易用map写出一个版本，使用map计数，返回数量为N的那个数

时间复杂度 (O(n))，n是A的size()

空间复杂度 (O(n))，n是A的size()

int repeatedNTimes(vector<int>& A) {
    map<int, int> _map;
    for(int val : A)
    {
        if (++_map[val] == (A.size()/2))
        {
            return val;
        }
    }
    return 0;
}

再进一步，要找到重复出现N次的那个数，由于其他数都是唯一的，只要找到重复出现2次的数就行了

这个比之前的版本快那么一点点，本质上差异并不大

时间复杂度 (O(n))，n是A的size()

空间复杂度 (O(n))，n是A的size()

int repeatedNTimes(vector<int>& A) {
    map<int, int> _map;
    for(int val : A)
    {
        if (++_map[val] == 2)
        {
            return val;
        }
    }
    return 0;
}

另一种办法是比较，从LeetCode上看到的

If we ever find a repeated element, it must be the answer. Let's call this answer the major element.

Consider all subarrays of length 4. There must be a major element in at least one such subarray.

This is because either:

• There is a major element in a length 2 subarray, or;
• Every length 2 subarray has exactly 1 major element, which means that a length 4 subarray that begins at a major element will have 2 major elements.

Thus, we only have to compare elements with their neighbors that are distance 1, 2, or 3 away.

只要找到一个重复的值，这个值就是要找的

至少有一个长度为4的子序列中存在重复的值

LeetCode的讨论区回答说需要检查的子序列的长度可以减小到3

时间复杂度 (O(n))，n是A的size()

空间复杂度 (O(1))

int repeatedNTimes(vector<int>& A) {
    for(int i = 0; i < A.size()-2; i++)
    {
        if(A[i] == A[i+1] || A[i] == A[i+2])
        {
            return A[i];
        }
    }
    return A[A.size()-1];
}

Reference

1.https://leetcode.com/problems/n-repeated-element-in-size-2n-array/solution/