list_need = [2,3,1,0,2,5,3]
list_t = []
for i in list_need:
if i not in list_t:
list_t.append(i)
else:
print(i)
按顺序计数，第一个计数大于1 的就返回
def find_number(li:list):
for i in li:
if li.count(i)>1:
return i
assert find_number([2, 3, 1, 0, 2, 5, 3])==2
list1 = [2,3,1,0,2,5,3]
list2 = []
for i in list1:
a = 0
for j in list1:
if i == j:
a += 1
while a > 1:
list2.append(a)
break
print(list2[0])
def firstrepeatnumber(lis: list) -> int:
for i in range(len(lis)):
po = lis.pop(0)
if po in lis:
return po
assert firstrepeatnumber([1,2,3,4]) == None
assert firstrepeatnumber([2,1,3,4,1,2,3,4]) == 2
assert firstrepeatnumber([1,2,3,4,4]) == 4
assert firstrepeatnumber([1,1,3,3]) == 1
assert firstrepeatnumber([1,2,2,4]) == 2
1 个赞
def listrep(li):
i, j = 0, 1
for i in range(len(li)-1):
for j in range(i, len(li)):
if i != j and li[i] == li[j]:
print(li[i])
break
listrep([1, 2, 3, 2])
list1 = [2, 3, 1, 0, 2, 5, 3]
for a in list1:
for b in list1:
if a == b:
print(b)
else:
break
def solution(nums):
flag = 1
while flag:
i = 0
while i < len(nums):
for j in range(i + 1, len(nums)):
if nums[i] == nums[j]:
flag = 0
return nums[i]
if flag == 1:
i += 1
assert solution([2,3,1,0,2,5,3]) == 2
assert solution([2,3,1,0,5,3]) == 3
import org.junit.Test;
import java.util.*;
import static org.junit.jupiter.api.Assertions.*;
public class RepeatedNumber {
@Test
public void testCase() {
assertAll(()->{
assertEquals(repeatedNumber(new int[]{1, 0, 3, 2, 3, 4, 6, 7, 7}),3);
assertEquals(repeatedNumber(new int[]{2,3,1,0,2,5,3}),2);
assertEquals(repeatedNumber(new int[]{ 2, 3, 4, 6, 7, 7}),7);
assertEquals(repeatedNumber(new int[]{1, 0, 3, 2, 4, 6, 7, }),-1);
});
}
public int repeatedNumber(int[] list) {
Set<Integer> map = new HashSet<>();
int repeat = -1;
for (int num : list) {
if (!map.add(num)) {
repeat = num;
break;
}
}
return repeat;
}
}
def solution(nums: list):
# your code here
return [i for i in set(nums) if nums.count(i) > 1][0]
assert solution([2, 3, 1, 0, 2, 5, 3]) == 2
assert solution([5, 2, 4, 3, 3, 2, 1]) == 2
def solution(nums: list):
from collections import Counter
for value,count in Counter(nums).items():
if count >1:
return value
#思路：每个元素和他前面的元素对比，相同就返回。不同继续找后一个
def find_firstdulp(nums):
for i in range(len(nums)-1):
if nums[i+1] in nums[0:i+1]:
return nums[i+1]
nums = [2,3,3,2]
print(find_firstdulp(nums))
def find_num(li):
for i in li:
if li.count(i) > 1:
return i
public int reSeeMember(int[] memberList) {
int reSeeMember = 0;
List list =new ArrayList<>();
for (int i = 0; i < memberList.length; i++) {
for (int j = i+1; j < memberList.length; j++) {
if (memberList[i] == memberList[j]) {
//使用赋值的形式后面的会覆盖前面的内容。所以使用集合
list.add(memberList[i]);
}
}
}
//取集合中的第一个元素，第一个为最开始重复的数
reSeeMember = (int) list.get(0);
return reSeeMember;
}
@Test
void testCase_03(){
assertAll(()->{
assertEquals(reSeeMember(new int []{2,3,1,0,2,5,3}),2);
assertEquals(reSeeMember(new int []{23,45,2,45,67,8}),45);
assertEquals(reSeeMember(new int []{21,31,11,11,22,50,39}),11);
assertEquals(reSeeMember(new int []{222,32,13,3,2,5,3}),3);
});
}
def findfirstrepeatnum(M:list):
for i in range(len(M) - 1):
for j in range(1 + i, len(M)):
if M[i] == M[j]:
return M[i]
break
findfirstrepeatnum([3,3,1,0,2,5,3])==3count=0
def fun(list1):
for i in list1:
global count
count=list1.count(i)
if count>1:
print(f’第一个重复的数据字是{i}，重复次数：{count}')
break
fun([2,3,1,0,2,5,3])
def solution(nums: list):
for i in range(len(nums)):
if nums[i] in nums[:i]:
return nums[i]
li = [2,3,1,0,2,5,3]
t = {}
for i in li:
if t.get(i):
print(i)
break
else:
t[i] = 1
lst = [6,8,5,3,4,2,1,7,3,5,8,9]
def repeatedNum(lst):
for i in range(len(lst)):
for j in range(i+1,len(lst)):
if lst[i] == lst[j]:
print("第一组相同的数为：", lst[i])
break
else:
continue
break
repeatedNum(lst)
def double_num(nums):
dict={}
for i in range(len(nums)):
if nums[i] not in dict.keys():
dict[nums[i]]=1
else:
return nums[i]
assert double_num([2,3,1,0,2,5,3])==2