给定一个数字列表和目标索引值，请编写一个函数，返回大于指定索引值对应元素的所有元素中，值最小的那个的索引值。如果不存在则返回-1。
示例：
输入： [4, 1, 3, 5, 6]，0
输出：3。
因为索引0对应的元素值是4，最小的大于4的数是5，5的索引是3，因此返回3。
题目难度：简单
题目来源：CodeWars：Least Larger
def solution(nums: list, i: int) -> int:
# your code here
assert solution([4, 1, 3, 5, 6], 0) == 3
assert solution([4, 1, 3, 5, 6], 4) == -1
assert solution([1, 3, 5, 2, 4], 0) == 3
def solution(nums: list, i: int) -> int:
# your code here
list1 = list(set(nums))
for j, k in enumerate(list1):
if k == nums[i]:
if j + 1 < len(list1):
num = list1[j + 1]
break
else:
return -1
if j == len(list1) - 1:
return -1
for j, k in enumerate(nums):
if k == num:
return j
def solution(nums: list, i: int) -> int:
num = nums[i]
# 如果num是列表中最大值，返回-1
if num == max(nums):
return -1
else:
# 列表升序排序，赋值新列表
nums_sort =sorted(nums)
# 新列表中获取大于num的最小值的索引，num的索引值+1
n_index = nums_sort.index(num)+1
# 新列表中获取大于num的最小值
result = nums_sort[n_index]
# 旧列表中返回大于num的最小值的索引
return nums.index(result)
assert solution([4, 1, 3, 5, 6], 0) == 3
assert solution([4, 1, 3, 5, 6], 4) == -1
assert solution([1, 3, 5, 2, 4], 0) == 3
def least_larger(a, i):
n = a[i]
min = n-1
min_index = -1
for i,num in enumerate(a):
if num > n and (min < n or num < min):
min = num
min_index = i
return min_indexdef solution(nums: list, i: int) -> int:
# your code here
list_num = []
for j in nums:
if j > nums[i]:
list_num.append(j)
if not list_num:
return -1
else:
return nums.index(min(list_num))
assert solution([4, 1, 3, 5, 6], 0) == 3
assert solution([4, 1, 3, 5, 6], 4) == -1
assert solution([1, 3, 5, 2, 4], 0) == 3def solution(nums: list, i: int) -> int:
# 0308 极小较大值
res = [x for x in nums if x > nums[i]]
if len(res) > 0 :
return nums.index(sorted(res)[0])
return -1
assert solution([4, 1, 3, 5, 6], 0) == 3
assert solution([4, 1, 3, 5, 6], 4) == -1
assert solution([1, 3, 5, 2, 4], 0) == 3
def solution(nums: list, i: int) -> int:
lis = nums[i+1:]
if lis == []:
return -1
else:
if max(lis) > nums[i]:
for p in range(len(lis)):
if lis[p] > nums[i]:
mi = lis[p]
ind = p
break
for q in range(len(lis)):
if lis[q] > nums[i] and lis[q] < mi:
mi = lis[q]
ind = q
return ind + 1 + i
else:
return -1
‘’’
【input】
nums: list
i:int
【output】
int
‘’’
def solution(nums, i):
datas = [x for x in nums if x > nums[i]]
if datas != :
return nums.index(min(datas))
else:
return -1
def solution(nums: list, i: int) -> int:
num = [j for j in nums if nums[i] < j]
if num != []:
return nums.index(min(num))
else:
return -1
assert solution([4, 1, 3, 5, 6], 0) == 3
assert solution([4, 1, 3, 5, 6], 4) == -1
assert solution([1, 3, 5, 2, 4], 0) == 3
def solution(nums: list, i: int) -> int:
return -1 if max(nums) == nums[i] else nums.index(min([j for j in nums if j > nums[i]]))
assert solution([4, 1, 3, 5, 6], 0) == 3
assert solution([4, 1, 3, 5, 6], 4) == -1
assert solution([1, 3, 5, 2, 4], 0) == 3
def solution(nums: list, i: int) -> int:
res = {}
if i >= len(nums) - 1:
return -1
for index,value in enumerate(nums):
if nums[index] > nums[i]:
res[nums[index]] = index
return res[min([key for key in res])]
assert solution([4, 1, 3, 5, 6], 0) == 3
assert solution([4, 1, 3, 5, 6], 4) == -1
assert solution([1, 3, 5, 2, 4], 0) == 3
def solution(nums: list, i: int) -> int:
target = nums[i]
result = dict()
if target == max(nums):
return -1
else:
for k,v in enumerate(nums):
if v > target and v not in result:
result[v] = k
return result[sorted(result.keys())[0]]
def miniMax(li, i):
n = li[i]
li1 = [j for j in li if n < j]
if li1 == :
return -1
else:
return li.index(min(li1))
assert miniMax([4, 1, 3, 5, 6], 0) == 3
assert miniMax([4, 1, 3, 5, 6], 4) == -1
assert miniMax([4, 1, 3, 8, 7, 6, 5], 0) == 6
def miniMax1(li, n):
m = li[n]
li1 =
for i in li:
if i > m:
li1.append(i)
if li1 == :
return -1
else:
return li.index(min(li1))
assert miniMax1([4, 1, 3, 5, 6], 0) == 3
assert miniMax1([4, 1, 3, 5, 6], 4) == -1
assert miniMax1([4, 1, 3, 8, 7, 6, 5], 0) == 6
def solution(nums: list, i: int) -> int:
if nums[i] == max(nums):
return -1
else:
a = nums[i]
ls = [x for x in nums if x >a ]
min_num =min(ls)
return nums.index(min_num)
如果列表中有重复的元素，这样写是有问题的
def solution(nums: list, i: int) -> int:
# your code here
if nums[i] == max(nums):
return -1
sort_nums = set(sorted(nums))
for idx,num in enumerate(sort_nums):
if num > nums[i]:
return nums.index(num)
assert solution([4, 1, 3, 5, 6], 0) == 3
assert solution([4, 1, 3, 5, 6], 4) == -1
assert solution([1, 3, 5, 2, 4], 0) == 3
def solution(nums: list, i: int) -> int:
res_list = [nums[j] for j in range(len(nums)) if nums[j] > nums[i]]
if res_list == []:
return -1
else:
res_list.sort()
return nums.index(res_list[0])
assert solution([4, 1, 3, 5, 6], 0) == 3
assert solution([4, 1, 3, 5, 6], 4) == -1
assert solution([1, 3, 5, 2, 4], 0) == 3