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 = 0;
        int temp = 0;
        
        for (int i = 0; i < nums.size(); ++i)
        {
            if (nums[i] == 1)
            {                                
                temp++;
                max = temp > max ? temp : max;
            }
            else
            {             
                temp = 0;
            }
        }
        
        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 = 0, max = 0;
        for (int n : nums)
            max = Math.max(max, maxHere = n == 0 ? 0 : maxHere + 1);
        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 = 0;
 int sum = 0;
 for (int i=0; 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..