class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        if(s.length()==0)
            return 0;
        vector<int> dp(s.size(),0);

        dp[0] = 1;   // 初始狀態。
        // dp[i] 包含當前字元的子串長度
        for(int i = 1; i < s.length(); i++){
            int k = -1;
            for(int
 
 j = 0; j <i; j++){
                if(s[i] == s[j])
                    k = j;
            }
            if(k>=0)
                // dp[i] = i-k;
                dp[i] = min(i-k,dp[i-1]+1);  // 針對abba這種情況
            else 
                dp[i] = dp[i-1] + 1;
            
        }
        int res = INT_MIN;
        
 
for(int i = 0; i < dp.size(); i++){
            res = max(res, dp[i]);
            cout<<"dp:"<<dp[i]<<endl;
        }
        return res;
    }
};

進一步優化

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        if(s.length()==0)
            return 0;
        vector
 
<int> dp(s.size(),0);

        dp[0] = 1;   // 初始狀態。
        // dp[i] 包含當前字元的子串長度
        int start = 0;
        for(int i = 1; i < s.length(); i++){
            int k = -1;
            for(int j = start; j <i; j++){
                if(s[i] == s[j]){
                    start = j;
                    k = j;
                }

            }
            if(k>=0)
                 dp[i] = i-k;
               // dp[i] = min(i-k,dp[i-1]+1);  // 針對abba這種情況
            else 
                dp[i] = dp[i-1] + 1;
            
        }
        int res = INT_MIN;
        for(int i = 0; i < dp.size(); i++){
            res = max(res, dp[i]);
            cout<<"dp:"<<dp[i]<<endl;
        }
        return res;
    }
};