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
这道题让我们求最大连续1的个数,不是一道难题。我们可以遍历一遍数组,用一个计数器cnt来统计1的个数,方法是如果当前数字为0,那么cnt重置为0,如果不是0,cnt自增1,然后每次更新结果res即可,参见代码如下:
解法一:
class Solution { public: int findMaxConsecutiveOnes(vector<int>& nums) { int res = 0, cnt = 0; for (int num : nums) { cnt = (num == 0) ? 0 : cnt + 1; res = max(res, cnt); } return res; } };
由于是个二进制数组,所以数组中的数字只能是0或1,那么连续1的和跟个数相等,所以我们可以计算和,通过加上num,再乘以num来计算,如果当前数字是0,那么sum就被重置为0,还是要更新结果res,参见代码如下:
解法二:
class Solution { public: int findMaxConsecutiveOnes(vector<int>& nums) { int res = 0, sum = 0; for (int num : nums) { sum = (sum + num) * num; res = max(res, sum); } return res; } };
本文转自博客园Grandyang的博客,原文链接:最大连续1的个数Max Consecutive Ones ,如需转载请自行联系原博主。