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

思路：找出数组中连续为1的个数最大值，遇到为0的值，遍历的i重置为0

优秀代码：

int findmax(vector<int>& nums){
int m = 0， MAX= 0；
for(auto n : nums){
if(n == 0){
MAX = max(MAX, m);
m = 0;
}
else
m++;
}
return max(MAx, m);
}

本人代码：（不简洁）

int findmax(vector<int>& nums){
int m = 0; max = 0;
for(size_t i = 0; i < nums.size(); i++){
if(nums[i] == 1)
m++;
if((i < nums.size() - 1 && nums[i+1] == 0) || i == nums.size() - 1){//当没有遍历到数组末尾，下一个为0，或者已经到了末尾，没有下一个
if(m > max)
max = m;
m = 0;
}
}
}

[Array]485. Max Consecutive Ones