1. 我自己写的

解法：如果有重复数字就直接输出最小的重复数字，如果没有就直接把两个列表里最小的数字拼在一起

（Python 3和Java的具体实现逻辑稍微有点差异，因为Java没有Python 3那么方便的API）

Python 3版：

class Solution:def minNumber(self, nums1: List[int], nums2: List[int]) -> int:#判断是否存在重复repetition_numbers=[]for number in nums1:if number in nums2:repetition_numbers.append(number)if len(repetition_numbers)>0:return min(repetition_numbers)else:nums1.sort()nums2.sort()if nums1[0]>nums2[0]:min_number=nums2[0]max_number=nums1[0]else:min_number=nums1[0]max_number=nums2[0]return int(str(min_number)+str(max_number))

Java版：

class Solution {public int minNumber(int[] nums1, int[] nums2) {int[] helper=new int[10];int min1=9,min2=9;for(int number:nums1){helper[number]+=1;min1=Math.min(min1,number);}for(int number:nums2){helper[number]+=1;min2=Math.min(min2,number);}//找重复值for(int number=1;number<=9;number++){if(helper[number]==2){return number;}}if(min1>min2){return min1+min2*10;}else{return min1*10+min2;}}
}

2. 题解

2.1 用哈希表来判断重复数字

class Solution {public int minNumber(int[] nums1, int[] nums2) {int s = same(nums1, nums2);if (s != -1) {return s;}int x = Arrays.stream(nums1).min().getAsInt();int y = Arrays.stream(nums2).min().getAsInt();return Math.min(x * 10 + y, y * 10 + x);}public int same(int[] nums1, int[] nums2) {Set<Integer> s = new HashSet<Integer>();for (int num : nums1) {s.add(num);}int x = 10;for (int num : nums2) {if (s.contains(num)) {x = Math.min(x, num);}}return x == 10 ? -1 : x;}
}

class Solution:def minNumber(self, nums1: List[int], nums2: List[int]) -> int:def same() -> int:s = set(nums1) & set(nums2)return -1 if not s else min(s)if (x := same()) != -1:return xx = min(nums1)y = min(nums2)return min(x * 10 + y, y * 10 + x)

2.2 用位运算判断重复数字

class Solution {public int minNumber(int[] nums1, int[] nums2) {int a = 0, b = 0;for (int num : nums1) a |= 1 << num;for (int num : nums2) b |= 1 << num;int c = a & b;if (c != 0)return Integer.numberOfTrailingZeros(c);int min1 = Integer.numberOfTrailingZeros(a);int min2 = Integer.numberOfTrailingZeros(b);return min1 < min2 ? min1 * 10 + min2 : min2 * 10 + min1;}
}

class Solution:def minNumber(self, nums1: List[int], nums2: List[int]) -> int:mask1 = mask2 = 0for x in nums1:mask1 |= 1 << xfor x in nums2:mask2 |= 1 << xmask = mask1 & mask2if mask:return (mask & -mask).bit_length() - 1a = (mask1 & -mask1).bit_length() - 1b = (mask2 & -mask2).bit_length() - 1return min(a * 10 + b, b * 10 + a)