1371. Find the Longest Substring Containing Vowels in Even Counts

如果按照正常思路，估计要o（N*N）的复杂度，采用累加法，只用O(N)的复杂度，只有这样才能通过

class Solution {
public:
    // 累加法，直接o1复杂度，但是存储空间会变大
    //存储aeiou 2^5=32种情况的所有index
    
    int convert(map<int, int>& cnt){
        int a = cnt['a']%2*pow(2, 0);
        int e = cnt['e']%2*pow(2, 1);
        int i = cnt['i']%2*pow(2, 2);
        int o = cnt['o']%2*pow(2, 3);
        int u = cnt['u']%2*pow(2, 4);
        return (a+e+i+o+u);
    }
    
    int findTheLongestSubstring(string s) {
        vector<int> temp;
        vector<vector<int>> res(32, temp);
        res[0].push_back(-1);
        
        map<int,int> cnt;
        for(int i=0;i<s.size();i++){
            if (cnt.find(s[i])==cnt.end()){
                cnt[s[i]] = 1;
            }
            else{
                cnt[s[i]] ++;
            }
            res[convert(cnt)].push_back(i);
        }
        
        int max = 0;
        for(int i=0;i<32;i++){
            if (res[i].size()==0){
                continue;
            }
            int a = res[i][0];
            int b = res[i][res[i].size()-1];
            max = max > abs(a-b)? max: abs(a-b);
        }
        return max;
    }
};

来源：CSDN

作者：zeroQiaoba

链接：https://blog.csdn.net/zeroQiaoba/article/details/104748419