[LeetCode]Student Attendance Record I

这道题比较简单，按照题目要求来即可，线性扫一遍，时间复杂度O(N)，空间复杂度O(1)，代码如下：

	class Solution {
	public:
	bool checkRecord(string s) {
	int len = s.size(), cntA = 0, i = 0;
	while(i < len)
	{
	char c = s[i];
	if(c == 'A')
	{
	if(++cntA > 1)return false;
	}
	else if(c == 'L')
	{
	int j = i;
	while(j < len && s[j] == 'L')++j;
	if(j - i > 2)return false;
	i = j;
	continue;
	}
	++i;
	}
	return true;
	}
	};

view raw Student Attendance Record I.cpp hosted with ❤ by GitHub