[leetcode-485-Max Consecutive Ones]

weixin_33719619

于 2017-05-12 11:09:00 发布

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原文链接：http://www.cnblogs.com/hellowooorld/p/6844605.html

本文介绍了一种通过遍历二进制数组来找到最长连续1序列的方法。算法使用一个计数器跟踪当前连续1的数量，并在遇到0时更新最大连续1的记录。此方法简单高效，适用于快速解决此类问题。

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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

思路：

遍历一遍，随时观察是否为1，不是的话从0开始。

int findMaxConsecutiveOnes(vector<int>& nums)
     {
         if (nums.size() == 0)return 0;
         int one = 0;
         int ret = 0;
         for (int i = 0; i < nums.size();i++)
         {
             if (nums[i] == 1) one++;            
             else if (nums[i] == 0)
             {
                 ret = max(ret, one);
                 one = 0;
             }
         }
         ret = max(ret, one);
         return ret;
     }