[LeetCode] 293. Flip Game 翻转游戏

You are playing the following Flip Game with your friend: Given a string that contains only these two characters: + and -, you and your friend take turns to flip twoconsecutive "++" into "--". The game ends when a person can no longer make a move and therefore the other person will be the winner.

Write a function to compute all possible states of the string after one valid move.

For example, given s = "++++", after one move, it may become one of the following states:

[
  "--++",
  "+--+",
  "++--"
]

If there is no valid move, return an empty list [].

给一个只含有'+', '-'的字符串，每次可翻动两个连续的'+'，求有多少种翻法。

解法：

java:

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public class Solution {     public List<String> generatePossibleNextMoves(String s) {         List<String> res = new ArrayList<>();         char[] arr = s.toCharArray();         for(int i = 1; i < s.length(); i++) {             if(arr[i] == '+' && arr[i - 1] == '+') {                 arr[i] = '-';                 arr[i - 1] = '-';                 res.add(String.valueOf(arr));                 arr[i] = '+';                 arr[i - 1] = '+';             }         }                   return res;     } }

Python:

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class Solution(object):     def generatePossibleNextMoves(self, s):         """         :type s: str         :rtype: List[str]         """         res = []         i, n = 0, len(s) - 1         while i < n:                                    # O(n) time             if s[i] == '+':                 while i < n and s[i+1] == '+':          # O(c) time                     res.append(s[:i] + '--' + s[i+2:])  # O(n) time and space                     i += 1             i += 1         return res

Python:

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# Time:  O(c * m * n + n) = O(c * n + n), where m = 2 in this question # Space: O(n) # This solution compares O(m) = O(2) times for two consecutive "+", where m is length of the pattern class Solution2(object):   def generatePossibleNextMoves(self, s):       """       :type s: str       :rtype: List[str]       """       return [s[:i] + "--" + s[i+2:] for i in xrange(len(s) - 1) if s[i:i+2] == "++"]

C++:

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// Time:  O(c * n + n) = O(n * (c+1)), n is length of string, c is count of "++"  // Space: O(1), no extra space excluding the result which requires at most O(n^2) space  class Solution {   public:       vector<string> generatePossibleNextMoves(string s) {           vector<string> res;           int n = s.length();           for (int i = 0; i < n - 1; ++i) {  // O(n) time               if (s[i] == '+') {                   for (;i < n - 1 && s[i + 1] == '+'; ++i) {  // O(c) time                       s[i] = s[i + 1] = '-';                       res.emplace_back(s);  // O(n) to copy a string                       s[i] = s[i + 1] = '+';                   }               }           }           return res;       }   };

C++:

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class Solution { public:     vector<string> generatePossibleNextMoves(string s) {         vector<string> res;         for (int i = 1; i < s.size(); ++i) {             if (s[i] == '+' && s[i - 1] == '+') {                 res.push_back(s.substr(0, i - 1) + "--" + s.substr(i + 1));             }         }         return res;     } };

　　

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