485. Max Consecutive Ones【easy】

485. Max Consecutive Ones【easy】

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

解法一：

 class Solution {
 public:
     int findMaxConsecutiveOnes(vector<int>& nums) {
         int max = ;
         int temp = ;
 
         for (int i = ; i < nums.size(); ++i)
         {
             if (nums[i] == )
             {
                 temp++;
                 max = temp > max ? temp : max;
             }
             else
             {
                 temp = ;
             }
         }
 
         return max;
     }
 };

思路很简单：是1就累加并且判断是否需要更新max，不是1就把累加和归为0，继续遍历。

 

解法二：

 public int findMaxConsecutiveOnes(int[] nums) {
         int maxHere = , max = ;
         for (int n : nums)
             max = Math.max(max, maxHere = n ==  ?  : maxHere + );
         return max;
     } 

大神解释如下：

The idea is to reset maxHere to 0 if we see 0, otherwise increase maxHere by 1
The max of all maxHere is the solution

110111
^ maxHere = 1
110111
.^ maxHere = 2
110111
..^ maxHere = 0
110111
...^ maxHere = 1
110111
....^ maxHere = 2
110111
.....^ maxHere = 3

解法三：

 int findMaxConsecutiveOnes(int* nums, int numsSize) {
  int max = ;
  int sum = ;
  for (int i=; i<numsSize; i++)
  {
      sum = (sum+nums[i])*nums[i];
      if(max<sum){max=sum;}
  }
 return max;
 }

这方法更牛逼，大神解释如下：Use the fact that multiplication with 0 resets everything..