剑指Offer：不修改数组找出重复的数字

题目来源：https://www.acwing.com/problem/content/description/15/
方法一：哈希表

class Solution {
public:
    int duplicateInArray(vector<int>& nums) 
    {
        unordered_map<int, int> umap;
        for (int i = 0; i < nums.size(); i ++ )
        {
            if (umap.find(nums[i]) == umap.end())
                umap.insert({nums[i], 1});
            else return nums[i];
        }
        return -1;
    }
};

方法二：vector代替哈希

class Solution {
public:
    int duplicateInArray(vector<int>& nums) 
    {
        vector<bool> result(nums.size(), false);
        for (int i = 0; i < nums.size(); i ++ )
        {
            if (result[nums[i]] == false)
                result[nums[i]] = true;
            else return nums[i];
        }
        return -1;
    }
};

方法三：下标法

class Solution {
public:
    int duplicateInArray(vector<int> nums) 
    {
        for (int i = 0; i < nums.size(); i ++ )
        {
            while (i != nums[i])
            {
                if (nums[i] == nums[nums[i]]) return nums[i];
                swap(nums[i], nums[nums[i]]);
            }
        }
        return -1;
    }
};