Python中偶數的最長序列

Question

使用非庫python代碼，我如何返回偶數最長序列的索引和計數？

a = [1, 3, 2, 6, 4, 1, 2, 2, 2, 8, 1] 

應該返回6和4，6是索引，4是計數。 我試過沒有運氣..

def evenSeq(list): 
    count=0 
    for i in list: 
     if list[i]%2 and list[i+1]%2==0: 
      count+=1 
    return count 

Answer 1

a=[1,3,2,6,4,2,2,2,2,2,1,2,2,2,8,1] 
def evenSeq(a): 
    largest = 0 
    temp_largest = 0 
    location = 0 
    for count, value in enumerate(a): 
     if value % 2 == 0: 
      temp_largest += 1 
     else: 
      temp_largest = 0 
     if temp_largest >= largest: 
      largest = temp_largest 
      location = count + 1 - temp_largest #plus one cause enumerate returns the index and we are subbing from the current streak which needs to be offset by one 
    return location, largest 

print(evenSeq(a)) #returns 2 8 

不是最漂亮的解決方案，但它可以幫助教你這是怎麼回事和解決方案的基本邏輯。基本上，它會檢查數字是否爲偶數，並保持存儲在temp_largest中的當前連勝數。檢查temp_largest是否是當時最大的已知條紋，並從枚舉更新索引。

編輯基於評論：

for count, value in enumerate(a): 

這條線基本上穿過名單，投入value和值count當前索引。 enumerate()基本上會經歷通過它的任何東西，並返回一個從0開始的計數。例如見下文。

a=[1,3,2,6,4,2,2,2,2,2,1,2,2,2,8,1] 
for index, value in enumerate(a): 
    print('{} index and value is {}'.format(index,value)) 

打印出：

0 index and value is 1 
1 index and value is 3 
2 index and value is 2 
3 index and value is 6 
4 index and value is 4 
5 index and value is 2 
6 index and value is 2 
7 index and value is 2 
8 index and value is 2 
9 index and value is 2 
10 index and value is 1 
11 index and value is 2 
12 index and value is 2 
13 index and value is 2 
14 index and value is 8 
15 index and value is 1 

Answer 2

這裏是一個可能的解決方案：

def even_seq(l): 
    best = (-1, -1) 
    start_i = 0 
    count = 0 
    for i, n in enumerate(l): 
     if n % 2 == 0: 
      count += 1 
      if count > best[1]: 
       best = (start_i, count) 
     else: 
      start_i = i + 1 
      count = 0 

    return best 

Answer 3

我會嘗試這種方式：

def evenSeq(seq): 
    i = startindex = maxindex = maxcount = 0 

    while i < len(seq): 
     if seq[i]%2==0: 
      startindex = i 
      while i < len(seq) and seq[i]%2==0: 
       i+=1 
      if maxcount < i - startindex: 
       maxcount = i - startindex 
       maxindex = startindex    
     i+=1 

    return (maxindex, maxcount)