[leetcode]5337. 每個元音包含偶數次的最長子字符串

前綴和，下次記住前綴和，從1開始存,cnt[0] 存0

class Solution {
public:
    int findTheLongestSubstring(string s) {
        int len = s.size();
        vector<int>cntA(len+1, 0);
        vector<int>cntE(len+1, 0);
        vector<int>cntI(len+1, 0);
        vector<int>cntO(len+1, 0);
        vector<int>cntU(len+1, 0);
                
        for(int i = 1; i <= len; i++)
        {
            cntA[i] = cntA[i-1] + (s[i-1] == 'a');
            cntE[i] = cntE[i-1] + (s[i-1] == 'e');
            cntI[i] = cntI[i-1] + (s[i-1] == 'i');
            cntO[i] = cntO[i-1] + (s[i-1] == 'o');
            cntU[i] = cntU[i-1] + (s[i-1] == 'u');
        }
        int res = 0;
        for(int i = len; i > 0; i--)
        {
            for(int j = 0; j < i; j++)
            {
                if(i - j  < res)
                {
                    break;
                }
                if((cntA[i] - cntA[j]) % 2 == 0 && 
                   (cntE[i] - cntE[j]) % 2 == 0 && 
                   (cntI[i] - cntI[j]) % 2 == 0 && 
                   (cntO[i] - cntO[j]) % 2 == 0 && 
                   (cntU[i] - cntU[j]) % 2 == 0   )
                {
                    // cout<<"----------"<<endl;
                    // cout<<i<<" "<<j<<endl;
                    // cout<<cntA[i]<<" "<<cntA[j]<<endl;
                    // cout<<cntE[i]<<" "<<cntE[j]<<endl;
                    // cout<<cntI[i]<<" "<<cntI[j]<<endl;
                    // cout<<cntO[i]<<" "<<cntO[j]<<endl;
                    // cout<<cntU[i]<<" "<<cntU[j]<<endl;
                    res = i - j;
                    // cout<<res<<endl;
                }
            }
        }
        return res;
    }
};

教訓：下面代碼未通過

class Solution {
public:
    int findTheLongestSubstring(string s) {
        int len = s.size();
        vector<int>cntA(len, 0);
        vector<int>cntE(len, 0);
        vector<int>cntI(len, 0);
        vector<int>cntO(len, 0);
        vector<int>cntU(len, 0);
        
        cntA[0] =  (s[0] == 'a');
        cntE[0] =  (s[0] == 'e');
        cntI[0] =  (s[0] == 'i');
        cntO[0] =  (s[0] == 'o');
        cntU[0] =  (s[0] == 'u');
        
        for(int i = 1; i < len; i++)
        {
            cntA[i] = cntA[i-1] + (s[i] == 'a');
            cntE[i] = cntE[i-1] + (s[i] == 'e');
            cntI[i] = cntI[i-1] + (s[i] == 'i');
            cntO[i] = cntO[i-1] + (s[i] == 'o');
            cntU[i] = cntU[i-1] + (s[i] == 'u');
        }
        int res = 0;
        for(int i = len-1; i >= 0; i--)
        {
            for(int j = 0; j < i; j++)
            {
                if(i - j + 1 < res)
                {
                    break;
                }
                if((cntA[i] - cntA[j]) % 2 == 0 && 
                   (cntE[i] - cntE[j]) % 2 == 0 && 
                   (cntI[i] - cntI[j]) % 2 == 0 && 
                   (cntO[i] - cntO[j]) % 2 == 0 && 
                   (cntU[i] - cntU[j]) % 2 == 0   )
                {

                    res = i - j + 1;
                }
            }
        }
        return res;
    }
};