Python 124ms DFS Solution

  • 0

    I used DFS to solve this problem . In each loop, both 1 center case(aba) and two center case (abba) are considered. Tell me if you have any improvement suggestions! Thanks

    class Solution:
            # @param {string} s
            # @return {string[][]}
            def partition_rec(self, result, oneResult,startIndex):
                for i in range(startIndex,len(oneResult)-1):
                    // 1 center case
                    if oneResult[i] == oneResult[i+1]:
                        anotherResult = oneResult[:i]+[oneResult[i]+oneResult[i+1]]+oneResult[i+2:]
                        result += [anotherResult]
                        self.partition_rec(result, anotherResult,i)
                    // 2 center case
                    if i>0 and oneResult[i-1] == oneResult[i+1]:
                        anotherResult = oneResult[:i-1]+[oneResult[i-1]+oneResult[i]+oneResult[i+1]]+oneResult[i+2:]
                        result += [anotherResult]
                        self.partition_rec(result, anotherResult,i-1)
            def partition(self, s):
                if len(s) == 0:
                    return []
                result = [] 
                oneResult = []
                for i in range(len(s)):
                    oneResult += [s[i]]
                result += [oneResult]
                self.partition_rec(result, result[0], 0)
                return result

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