[Leetcode]485. Max Consecutive Ones

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：如果一开始传进来数组为空，则返回0。否则，设置两个变量maxConsecutiveOnes 和consecutiveOnes，

consecutiveOnes统计连续 1 的个数，如果某个位置导致连续中断，则把consecutiveOnes 赋0，继续统计后面的。

这个过程中把最大的consecutiveOnes值赋给maxConsecutiveOnes，代码如下

 1 class Solution {
 2     public int findMaxConsecutiveOnes(int[] nums) {
 3         if (nums.length==0)
 4             return 0;
 5         int consecutiveOnes = 0,maxConsecutiveOnes = 0;
 6         for (int i=0;i<nums.length;i++){
 7             if (nums[i]==1)
 8                 consecutiveOnes++;
 9             else 
10                 consecutiveOnes=0;
11             if (consecutiveOnes>maxConsecutiveOnes)
12                 maxConsecutiveOnes = consecutiveOnes;
13         }
14         return maxConsecutiveOnes;
15     }
16 }

[Leetcode]485. Max Consecutive Ones

原文：http://www.cnblogs.com/David-Lin/p/7742681.html