Solution

• 利用数组来保存出现次数

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param numbers int整型vector 
     * @return int整型
     */
    int duplicate(vector<int>& numbers) {
        // write code here
        int n = numbers.size();
        if (n < 2) return -1;
        vector<int> record(n, 0);
        for (int num : numbers) {
            record[num]++;
            if (record[num] > 1) return num;
        }
        return -1;
    }
};

• 利用数字范围条件，空间复杂度O(1)

class Solution {
public:
    int duplicate(vector<int>& numbers) {
        // write code here
        if (numbers.size() < 2) return -1;
        for (int i = 0; i < numbers.size(); ++i) {
            while (i != numbers[i]) {
                if (numbers[i] == numbers[numbers[i]])
                    return numbers[i];
                // 每个数放到它的索引位置
                swap(numbers[i], numbers[numbers[i]]);
            }
        }
        return -1;
    }
};