485. Max Consecutive Ones
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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.
这道题我主要是开了一个数组cnt来记录每一个对应的数组的位置的前面几个连续为1的数的和,然后再在当遇到0的时候去更新最大值maxx。这里要注意的是【0,1】这种情况,在循环里会少了这样一个判断,所以在跳出循环时,还得做一个maxx的更新。
代码如下:
class Solution {public: int findMaxConsecutiveOnes(vector<int>& nums) { vector<int>cnt(nums.size()); int maxx=0; cnt[0]=nums[0]; for(int i=0;i<nums.size();i++){ if(nums[i]==0){ cnt[i]=0; maxx=max(maxx,cnt[i-1]); } else{ cnt[i]=cnt[i-1]+1; } } maxx=max(maxx,cnt[nums.size()-1]); return maxx; }};
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- 485. Max Consecutive Ones*
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485. Max Consecutive Ones
- 485.Max Consecutive Ones
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