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每个元音包含偶数次的最长子字符串

解题思路

通过这道题,主要了解到了c++的位运算。包括了~ ,^.
一开始没有理解~运算,其实是位取反。
~-1 = 0

代码

cpp
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class Solution {
public:
int findTheLongestSubstring(string s) {
int ans=0;
int status=0;
vector<int> pos(1<<5, -1);
pos[0] = 0;
for(int i=0;i <s.size();i++){
if(s[i]=='a'){
status ^= 1<<0;
}else if(s[i]=='e'){
status ^= 1<<1;
}else if(s[i]=='i'){
status ^= 1<<2;
}else if(s[i]=='o'){
status ^= 1<<3;
}else if(s[i]=='u'){
status ^= 1<<4;
}
if(~pos[status]){
ans = max(ans, i+1-pos[status]);
}else{
pos[status] = i+1;
}
}
return ans;
}
};
Author: kim yhow
Link: http://yoursite.com/2020/05/20/每个元音包含偶数次的最长子字符串/
Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.
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