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.

Note:

• The input array will only contain 0 and 1.
• The length of input array is a positive integer and will not exceed 10,000

解法一：

 class Solution {

 public:

     int findMaxConsecutiveOnes(vector<int>& nums) {

         int max = ;

         int temp = ;

         for (int i = ; i < nums.size(); ++i)

         {

             if (nums[i] == )

             {

                 temp++;

                 max = temp > max ? temp : max;

             }

             else

             {

                 temp = ;

             }

         }

         return max;

     }

 };

思路很简单：是1就累加并且判断是否需要更新max，不是1就把累加和归为0，继续遍历。

 public int findMaxConsecutiveOnes(int[] nums) {

         int maxHere = , max = ;

         for (int n : nums)

             max = Math.max(max, maxHere = n ==  ?  : maxHere + );

         return max;

     } 

 int findMaxConsecutiveOnes(int* nums, int numsSize) {

  int max = ;

  int sum = ;

  for (int i=; i<numsSize; i++)

  {

      sum = (sum+nums[i])*nums[i];

      if(max<sum){max=sum;}

  }

 return max;

 }