8-lines C++, 7-lines Python


  • 5

    The idea is quite straightforward: just traverse s and each time when we see two consecutive +s, convert them to -s and add the resulting string to the final result moves. But remember to recover the string after that.

    The C++ code is as follows.

    class Solution {
    public:
        vector<string> generatePossibleNextMoves(string s) {
            vector<string> moves;
            int n = s.length();
            for (int i = 0; i < n - 1; i++) {
                if (s[i] == '+' && s[i + 1] == '+') { 
                    s[i] = s[i + 1] = '-';
                    moves.push_back(s);
                    s[i] = s[i + 1] = '+';
                }
            }
            return moves;
        }
    };
    

    Well I also try to write a Python solution since Python supports sequential comparisons, which is quite convenient. But Python does not support modifying a string and I can only use list and join to do the same thing.

    class Solution(object):
        def generatePossibleNextMoves(self, s):
            """
            :type s: str
            :rtype: List[str]
            """
            moves, n, s = [], len(s), list(s)
            for i in xrange(n - 1):
                if s[i] == s[i + 1] == '+': 
                    s[i] = s[i + 1] = '-'
                    moves += ''.join(s),
                    s[i] = s[i + 1] = '+' 
            return moves

  • 0
    T

    Hello,

    I have the same solution for C++. Any idea why it is just 7.5% above others in terns of run time ?


Log in to reply
 

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.