485. Max Consecutive Ones

最新推荐文章于 2018-06-01 17:10:23 发布

原创最新推荐文章于 2018-06-01 17:10:23 发布 · 338 阅读

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本文介绍了一种高效算法，用于找出二进制数组中最长的连续1序列。通过一次遍历，利用计数器跟踪当前连续1的数量，并更新最大值。

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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

这道题的做法是遍历一遍vector数组，设一个整数count来计数连续的1有多少个，遇到0的时候令count = 0，设一个整数Max保存count的最大值

代码如下：

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int Max = 0;
  int count = 0;
  for (int i = 0; i < nums.size(); i++){
   if (nums[i] == 1)
       count++;
   else
       count = 0;
   if (Max < count)
     Max = count;
  }
  return Max;
    }
};