485. 最大连续 1 的个数
1 class Solution:
2 def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
3 maxCount = count = 0
4
5 for i, num in enumerate(nums):
6 if num == 1:
7 count += 1
8 else:
9 maxCount = max(maxCount, count)
10 count = 0
11
12 maxCount = max(maxCount, count)
13 return maxCount
count = 0
ans = 0
for i in nums:
if i == 1:
count += 1
if ans < count:
ans = count
else:
count = 0
return ans
#先化为str,再用split方法将其以"0"分隔，结果便是list中最长的item的长度
s = ""
for i in nums:
s += str(i)
L2 = s.split("0")
return max([len(i) for i in L2])
#快慢同向双指针（移动窗口法）
slow = fast = 0
n = len(nums)
max_count = 0
while slow < n:
# 找出连续1中，第一个1的位置
if nums[slow] != 1:
slow += 1
else:
# fast指针从slow指针开始遍历，直至nums[fast]!=1
fast = slow
while fast < n and nums[fast] == 1:
fast += 1
# 局部最长fast-slow和全局最长max_count取最大值
max_count = max(max_count, fast - slow)
# 更新slow指针的位置到fast指针处，继续循环，寻找下一个1
slow = fast
return max_count
283. 移动零
1 class Solution:
2 def moveZeroes(self, nums: List[int]) -> None:
3 """
4 Do not return anything, modify nums in-place instead.
5 """
6 #不交换，之间覆盖
7 index = 0
8 i = 0
9 n = len(nums)
10 while i < n :
11 if nums[i] != 0:
12 nums[index] = nums[i]
13 index += 1
14 i += 1
15
16 #python 在 for 里修改索引变量不会影响到下一个循环，使用 while 才行
17 while index < n:
18 nums[index] = 0
19 index += 1
20
21 return nums
27. 移除元素
1 class Solution:
2 def removeElement(self, nums: List[int], val: int) -> int:
3 #pop删除元素
4 n = len(nums)
5 i = 0
6 if n == 0:
7 return 0
8
9 while i < n:
10 if nums[i] == val:
11 nums.pop(i)
12 n -= 1 #数组长度发生变化
13 else:
14 i += 1 #不等于val值才往下走一位
15
16 return len(nums)
17
18 #双指针 更快
19 s = f = 0
20
21 while f < len(nums):
22 if nums[f] != val:
23 nums[s] = nums[f]
24 s += 1
25 f += 1
26
27 return s
