static int wing=[]()
{
std::ios::sync_with_stdio(false);
cin.tie(NULL);
return ;
}(); class Solution
{
public:
bool isOneBitCharacter(vector<int>& bits)
{
int sz=bits.size();
int i=;
while(i<sz-)
{
if(bits[i]==)
i+=;
else
i+=;
}
return i==sz-;
}
};

这个题,从前向后扫,只要当前位为1,则不用管下一位是啥,必然要构成2位,所以当前位为1时,直接跳至下下个位置。

只有当末尾0前面有个落单的1时,末尾才不能为单,此时扫描到那个1的时候,加上2,i变为了sz

当末尾0前面不为非规则数字,i 必然不会跳过最后的0,会指向那个0,即i==sz-1

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