Given a binary array nums
, return the maximum number of consecutive 1
's in the array.
Example 1:
Input: nums = [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.
Example 2:
Input: nums = [1,0,1,1,0,1] Output: 2
Constraints:
1 <= nums.length <= 105
nums[i]
is either0
or1
.
class Solution:
def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
res = t = 0
for num in nums:
if num == 1:
t += 1
else:
res = max(res, t)
t = 0
return max(res, t)
class Solution {
public int findMaxConsecutiveOnes(int[] nums) {
int res = 0, t = 0;
for (int num : nums) {
if (num == 1) {
++t;
} else {
res = Math.max(res, t);
t = 0;
}
}
return Math.max(res, t);
}
}
/**
* @param {number[]} nums
* @return {number}
*/
var findMaxConsecutiveOnes = function (nums) {
let res = 0,
t = 0;
for (let num of nums) {
if (num == 1) {
++t;
} else {
res = Math.max(res, t);
t = 0;
}
}
return Math.max(res, t);
};