Leetcode question brushing record ｜ 485_ Maximum number of consecutive ones

---

subject

Given a binary array nums , Calculate the maximum continuity 1 The number of .

Example 1:

 Input ：nums = [1,1,0,1,1,1]
 Output ：3
 explain ： The first two digits and the last three digits are consecutive  1 , So the greatest continuity  1  The number of is  3.

Example 2:

 Input ：nums = [1,0,1,1,0,1]
 Output ：2

Tips ：

• 1 <= nums.length <= 105
• nums[i] No 0 Namely 1.

Their thinking

Train of thought reminder ： Double pointer （ Fast pointer and slow pointer ）
Train of thought details ：

1. Define two pointers , for example fast and slow（fast It is used to record several consecutive 1,slow It is used to record the number of final output ）
2. Place both pointers at the beginning of the array , That is to say, make fast = 0,slow = 0
3. The fast pointer traverses the entire array in turn , Every forward position , Value of fast pointer +1, When to take action ？
4. When the fast pointer encounters be equal to 0 When the number of , Take action at this time ： Store the value of the slow pointer as the maximum value between the fast pointer and the slow pointer at this time slow = max(slow,fast); Then set the value of the fast pointer again 0.
5. When the cycle is completed , Store the value of the slow pointer as the maximum value between the fast pointer and the slow pointer at this time slow = max(slow,fast);
6. Finally back to slow

Code implementation

c++

class Solution {
    
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
    
        int fast=0,slow = 0;
        int n = nums.size();
        for(int i = 0;i<n;i++){
    
            if(nums[i]==1){
    
                fast++;
            }
            else{
    
                slow = max(slow,fast);
                fast = 0;
            }
        }
        slow = max(slow,fast);
        return slow;
    }
};

python

class Solution:
    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
        if nums is None or len(nums)==0:
            return 0
        fast = 0
        slow = 0
        for i in range (0,len(nums)):
            if(nums[i]==1):
                fast =  fast+1
            else:
                slow = max(fast,slow)
                fast = 0
        slow = max(fast,slow)
        return slow