[LeetCode] 485. Max Consecutive Ones
grandyang opened this issue · comments
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
这道题让求最大连续1的个数,不是一道难题。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;
}
};
Github 同步地址:
类似题目:
参考资料:
https://leetcode.com/problems/max-consecutive-ones/