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Easy
String

中文文档

Description

You are playing a Flip Game with your friend.

You are given a string currentState that contains only '+' and '-'. You and your friend take turns to flip two consecutive "++" into "--". The game ends when a person can no longer make a move, and therefore the other person will be the winner.

Return all possible states of the string currentState after one valid move. You may return the answer in any order. If there is no valid move, return an empty list [].

 

Example 1:

Input: currentState = "++++"
Output: ["--++","+--+","++--"]

Example 2:

Input: currentState = "+"
Output: []

 

Constraints:

  • 1 <= currentState.length <= 500
  • currentState[i] is either '+' or '-'.

Solutions

Solution 1: Traversal + Simulation

We traverse the string. If the current character and the next character are both +, we change these two characters to -, add the result to the result array, and then change these two characters back to +.

After the traversal ends, we return the result array.

The time complexity is $O(n^2)$, where $n$ is the length of the string. Ignoring the space complexity of the result array, the space complexity is $O(n)$ or $O(1)$.

Python3

class Solution:
    def generatePossibleNextMoves(self, currentState: str) -> List[str]:
        s = list(currentState)
        ans = []
        for i, (a, b) in enumerate(pairwise(s)):
            if a == b == "+":
                s[i] = s[i + 1] = "-"
                ans.append("".join(s))
                s[i] = s[i + 1] = "+"
        return ans

Java

class Solution {
    public List<String> generatePossibleNextMoves(String currentState) {
        List<String> ans = new ArrayList<>();
        char[] s = currentState.toCharArray();
        for (int i = 0; i < s.length - 1; ++i) {
            if (s[i] == '+' && s[i + 1] == '+') {
                s[i] = '-';
                s[i + 1] = '-';
                ans.add(new String(s));
                s[i] = '+';
                s[i + 1] = '+';
            }
        }
        return ans;
    }
}

C++

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

Go

func generatePossibleNextMoves(currentState string) (ans []string) {
	s := []byte(currentState)
	for i := 0; i < len(s)-1; i++ {
		if s[i] == '+' && s[i+1] == '+' {
			s[i], s[i+1] = '-', '-'
			ans = append(ans, string(s))
			s[i], s[i+1] = '+', '+'
		}
	}
	return
}

TypeScript

function generatePossibleNextMoves(currentState: string): string[] {
    const s = currentState.split('');
    const ans: string[] = [];
    for (let i = 0; i < s.length - 1; ++i) {
        if (s[i] === '+' && s[i + 1] === '+') {
            s[i] = s[i + 1] = '-';
            ans.push(s.join(''));
            s[i] = s[i + 1] = '+';
        }
    }
    return ans;
}