3.配列内の繰り返し番号（findRepeatNumber）

3.配列内の繰り返し番号（findRepeatNumber）

1. python（1）

class Solution:
    def findRepeatNumber(self, nums: List[int]) -> int:
        sorted_nums = sorted(nums)
        sorted_nums_length = len(sorted_nums)
        for i in range(sorted_nums_length):
            if sorted_nums[i] == sorted_nums[i-1]:
                return sorted_nums[i]
        return -1

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2. python（2）

class Solution:
    def findRepeatNumber(self, nums: List[int]) -> int:
        dic = set()
        for num in nums:
            if num in dic:
                return num
                dic.add(num)
        return -1

3. Javaの

class Solution {
    
    
    public int findRepeatNumber(int[] nums) {
    
    
        Set<Integer> set = new HashSet<Integer>();
        int repeat =-1;
        for(int num:nums){
    
    
            if(!set.add(num)){
    
    
                repeat = num;
                break;
            }
        }
        return repeat;
    }
}

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