已知列表里有一组元素
[1, 2.5, 6, 6, 7, 7, 7, 2,5, 12]，请编写一个python3函数，找出列表中连续出现次数最多的数字及起始下标。题目难度：中等
题目来源：面试真题
贡献者：玉栋同学
def most_feq(nums:list) -> (int, int):
pass
assert most_feq([1, 2.5, 6, 6, 7, 7, 7, 2,5, 12]) == (7, 4)
from itertools import groupby
def most_feq(nums: list) -> (int, int, int):
max_time = max(len(list(v)) for k, v in groupby(nums)) #最大次数
max_index=1 #最大次数的起始位置
for index, (k, v) in enumerate(groupby(nums)):
if len(list(v)) == max_time:
max_index+= index
return k,max_index
assert most_feq([1, 2.5, 6, 6, 7, 7, 7, 2,5, 12]) == (7, 4)
def most_feq(nums: list):
max_time = 0 # 最大连续出现次数
continuous_time = 1 # 第几次连续出现
pre_element = None # 记录上一个元素
max_time_num = None # 最大次数的值
for index, i in enumerate(nums):
if i == pre_element:
max_time_num = i
continuous_time += 1
max_time = max(continuous_time, max_time)
else:
pre_element = i
continuous_time = 1
return max_time_num, max_time_num - max_time
assert most_feq([1, 2.5, 6, 6, 7, 7, 7, 2, 5, 12]) == (7, 4)
def most_feq(nums: list) -> (int, int):
temp_dict = {}
for n in nums:
if n in temp_dict:
temp_dict[n] += 1
else:
temp_dict[n] = 1
max_k = max(temp_dict, key=temp_dict.get)
return max_k, nums.index(max_k)
assert most_feq([1, 2.5, 6, 6, 7, 7, 7, 2, 5, 12]) == (7, 4)试下
assert most_feq([1, 1,2.5, 6, 6, 7, 7, 7, 2, 5, 12]) == (7, 4)
max_index有点看不懂为啥以1开始，为啥加index，我改了一下，你看下哈
max_time = max(len(list(v)) for k, v in groupby(nums)) # 最大次数
max_index = 0 # 最大次数的起始位置
for index, (k, v) in enumerate(groupby(nums)):
len_arr: int = len(list(v))
if len_arr == max_time:
return k, max_index
else:
max_index += len_arr
assert most_feq([1, 2, 6, 6, 7, 7, 7, 2, 1, 12]) == (7, 4)
assert most_feq([1, 2.5, 6, 6, 6, 7, 7, 2, 7, 7, 5, 12]) == (6, 2)
def most_feq(nums:list) -> (int, int):
if nums is None or len(nums) == 0:
return None, -1
max_item: int = nums[0]
max_count: int = 1
last_item: int = nums[0]
last_count: int = 1
for i in range(1, len(nums)):
if last_item == nums[i]:
last_count += 1
if last_count > max_count:
max_item = nums[i]
max_count = last_count
else:
last_item = nums[i]
last_count = 1
return max_item, nums.index(max_item)
def test_most_feq():
assert most_feq([1, 2, 6, 6, 7, 7, 7, 2, 1, 12]) == (7, 4)
assert most_feq([1, 2.5, 6, 6, 6, 7, 7, 2, 7, 7, 5, 12]) == (6, 2)
试试运行下这个
assert most_feq([1,1, 2, 6, 6, 7, 7, 7, 2, 1, 12]) == (7, 4)
max_index=1 起始位置就是从nums第一个数起
你这个断言是不是错了，应该断言(7, 5)吧？前面加了一位呀
我之前的没做连续判断，修改了一下
def most_feq(nums: list) -> (int, int):
index = []
count = 1
for x in range(len(nums)):
if x == len(nums) - 1:
index.append((x + 1 - count, count))
elif nums[x] == nums[x+1]:
count += 1
else:
index.append((x+1-count, count))
count = 1
max_index, _ = max(index, key=lambda y: y[1])
return nums[max_index], max_index
assert most_feq([1, 5, 2.5, 5, 6, 6, 7, 7, 7, 2, 5, 5]) == (7, 6)
assert most_feq([1, 1, 2.5, 6, 6, 7, 7, 7, 2, 5, 12]) == (7, 5)
public ArrayList<Integer> mostFeq(ArrayList<Integer> arrayList) {
// todo: 已知列表里有一组元素[1, 2，5, 6, 6, 7, 7, 7, 2,5, 12]，请编写一个python3函数，找出列表中连续出现次数最多的数字及起始下标。
int start = 0;
int end = 1;
int len = arrayList.size();
Map<String, ArrayList<Integer>> map = new HashMap<>();
int step = 0;
int max = 0;
while (end < len) {
if (arrayList.get(end).equals(arrayList.get(start))) {
step++;
end++;
} else {
ArrayList<Integer> integers1 = new ArrayList<>();
integers1.add(arrayList.get(start));
integers1.add(start);
integers1.add(start + step);
if (end - start + 1 > max) {
map.put("max", integers1);
max = end - start + 1;
}
start = end;
end++;
step = 0;
}
}
ArrayList<Integer> integers2 = new ArrayList<>();
integers2.add(arrayList.get(arrayList.size() - 1));
integers2.add(start);
integers2.add(end);
if (end - start + 1 > max) {
map.put("max", integers2);
}
return map.get("max");
}
@Test
public void mostFeqTest() {
ArrayList<Integer> list = new ArrayList<>();
list.add(56);
list.add(56);
list.add(56);
list.add(56);
list.add(56);
list.add(56);
list.add(56);
list.add(56);
list.add(56);
list.add(74);
list.add(99);
list.add(99);
list.add(99);
list.add(86);
list.add(180);
list.add(181);
list.add(181);
list.add(181);
list.add(181);
list.add(181);
list.add(181);
list.add(181);
ArrayList<Integer> integers = mostFeq(list);
}
def most_feq(nums:list) -> (int, int):
temp_nums=set(nums)
# 假设一个最大的数
max_number = nums[0]
# 假设次数
count = 0
for i in temp_nums:
temp=nums.count(i)
if temp > count:
count = temp
max_number= i
return (max_number, nums.index(max_number))
def most_feq(nums:list) -> (int, int):
""" 找出列表中连续出现次数最多的数字及起始下标"""
# nums = [1, 2.5, 6, 6, 7, 7, 7, 2, 5, 12]
item = 1
dic = {}
for i in range(0,len(nums)-1):
j = i+1
if nums[i] != nums[j]:
dic[nums[i]] = item
else:
j += 1
item += 1
max_nums = max(dic, key=dic.get)
max_index = nums.index(max_nums)
return max_nums,max_index
assert most_feq([1, 2.5, 6, 6, 7, 7, 7, 2, 5, 12]) == (7, 4)
def most_feq(nums: list) -> (int, int):
max_num = 0 # 最大的数
con_time = 0 # 连续次数
max_time = 0 # 最大的连续次数
max_index = 0 # 最大连续次数的元素下标
pro_num = 0 # 上一个元素
for i in range(0, len(nums) - 1):
if pro_num == nums[i]:
con_time += 1
if con_time >= max_time:
max_num = nums[i]
max_time = con_time
if con_time >= 2: # 连续次数大于2,最大值的下标=当前值的下标-连续次数
max_index = i - con_time
elif max_time == 1: # 少于2次的直接拿上一个元素的下标
max_index = i - 1
else:
pro_num = nums[i]
con_time = 0
return max_num, max_index
assert most_feq([1, 2.5, 6, 6, 7, 7, 7, 2, 5, 12]) == (7, 4)
assert most_feq([1, 2.5, 6, 6, 6, 7, 7, 7, 7, 7, 2, 2, 2, 2, 2, 2, 5, 12]) == (2, 10)
assert most_feq([1, 2.5, 6, 6, 7, 7, 2, 2, 2, 8, 8, 8, 8, 8, 5, 12]) == (8, 9)
assert most_feq([1, 1, 2.5, 6, 6, 8, 8, 8, 2, 5, 12]) == (8, 5)
assert most_feq([1, 0.2, 4, 4, 6, 7, 8]) == (4, 2)