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..

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