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

找出一个二进制串中连续的1的个数。

 1     public int findMaxConsecutiveOnes(int[] nums) {
 2          int max=0,i=0;
 3         while (i<nums.length)
 4         {
 5             int sum=0;
 6              while (i<nums.length && nums[i]==1)
 7             {
 8                 sum++;
 9                 i++;
10             }
11             i++;
12             max = Math.max(sum,max);
13         }
14         return max;       
15     }
原文地址:https://www.cnblogs.com/wzj4858/p/7669974.html