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
and1
. - The length of input array is a positive integer and will not exceed 10,000
题目标签:Array
题目给了我们一个nums array, 里面只会出现0和1, 让我们找到一个最长连续1的长度。
这题很简单,只要维护一个maxCount 就可以了,再设一个count,遇到1的时候,count++;遇到0的时候,把count 和 maxCount 中挑大的继续保存,更新count = 0。
注意,最后遍历完,如果最后一段是1的话,还需要维护maxCount一次
Java Solution:
Runtime beats 70.20%
完成日期:05/10/2017
关键词:Array
关键点:维护一个maxCount
public class Solution
{
public int findMaxConsecutiveOnes(int[] nums)
{
int maxCount = 0;
int count = 0; for(int i=0; i<nums.length; i++)
{
if(nums[i] == 1) // count consecutive ones
count++;
else if(nums[i] == 0) // if reach 0, save large count into maxCount
{
maxCount = Math.max(maxCount, count);
count = 0; // update count to 0
} }
// check the last consecutive ones
maxCount = Math.max(maxCount, count); return maxCount;
}
}
参考资料:N/A
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