Sword finger offer 03 Duplicate number in array

Original link ： The finger of the sword Offer 03. Repeated numbers in an array

 

solution:

        The hash table stores the numbers that have occurred , Time complexity O(n)

class Solution {
public:
    int findRepeatNumber(vector<int>& nums) {
       unordered_set<int> map;
       for(auto &x : nums) {
           if(map.count(x)) return x;
           map.insert(x);
       }
       return 0;         
    }
};

        Sort + Traverse , Time complexity O(nlogn)

class Solution {
public:
    int findRepeatNumber(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        for(int i = 1;i < nums.size();i++) {
            if(nums[i] == nums[i - 1]) return nums[i];
        }
        return 0;
    }
};

        Given element must be in 0~n - 1, If the number of occurrences is 1, It must be able to switch to the correct position .

class Solution {
public:
    int findRepeatNumber(vector<int>& nums) {
        for(int i = 0;i < nums.size();i++) {
            while(nums[i] != i) {
                if(nums[nums[i]] == nums[i]) return nums[i];
                swap(nums[nums[i]], nums[i]);
            }
        }
        return 0;
    }
};