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

这道题让求最大连续1的个数，不是一道难题。可以遍历一遍数组，用一个计数器 cnt 来统计1的个数，方法是如果当前数字为0，那么 cnt 重置为0，如果不是0，cnt 自增1，然后每次更新结果 res 即可，参见代码如下：

解法一：

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int res = 0, cnt = 0;
        for (int num : nums) {
            cnt = (num == 0) ? 0 : cnt + 1;
            res = max(res, cnt);
        }
        return res;
    }
};

由于是个二进制数组，所以数组中的数字只能是0或1，那么连续1的和跟个数相等，所以可以计算和，通过加上 num，再乘以 num 来计算，如果当前数字是0，那么 sum 就被重置为0，还是要更新结果 res，参见代码如下：

解法二：

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int res = 0, sum = 0;
        for (int num : nums) {
            sum = (sum + num) * num;
            res = max(res, sum);
        }
        return res;
    }
};

Github 同步地址：

#485

类似题目：

Max Consecutive Ones II

Max Consecutive Ones III

参考资料：

https://leetcode.com/problems/max-consecutive-ones/

https://leetcode.com/problems/max-consecutive-ones/discuss/96693/Java-4-lines-concise-solution-with-explanation

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