Given a binary array, find the maximum number of consecutive 1s in this array.

Example 1:

Input: [1,1,0,1,1,1]
Output: 3
Explanation: The first two digits or the last three digits are consecutive 1s.
    The maximum number of consecutive 1s is 3.

My idea:for循环遍历，最后退出循环的时候也要判断，因为为1的时候我不能去给max赋值

class Solution:
    def findMaxConsecutiveOnes(self, nums):
        count=0
        max=0
        for i in nums[::]:
            if(i==1):
                count=count+1
            if(i==0):
                if(max<count):
                    max=count
                count=0
        if (max < count):
            max = count
        return max

 

那用双指针怎么做呢，参考一下别人的，主要就是从slow开始数，fast往前跑，跑完一个循环给slow，就这样迭代

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int max = 0;
        
        int slow = 0;
        int fast = 0;
        int end = nums.length;
        
        while (slow < end) {
            fast = slow;
            int count = 0;
            while (fast < end && nums[fast++] == 1) {
                count++;
            }
            max = Math.max(max,count);
            slow = fast;
        }
        
        return max;
    }
}

ok 我试着用pyhton写一遍

class Solution:
    def findMaxConsecutiveOnes(self, nums):

        m = 0
        a = 0
        b = 0
        end = len(nums)
        while (a < end):
            b = a
            count = 0
            while (nums[b] == 1):
                count = count + 1
                b = b + 1
                if(b==end):
                    break
            m = max(m, count)
            a = b
        return m

ok,超出时间限制555