题解 | 字符串的排列

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param str string字符串 
# @return string字符串一维数组
#   参考： https://blog.csdn.net/YouMing_Li/article/details/114521466
class Solution:
    # 超时 
    def Permutation1(self , str: str) -> List[str]:
        # write code here
        #字符串转为数组
        n=len(str)
        arr=list(str)
        res=[]
        def backtrack(first=0):
            if n ==first:
                temp=arr.copy()
                st="".join(temp)
                if st not in res:
                    res.append(st)
            for i in range(first,n):
                arr[first],arr[i]=arr[i],arr[first]
                backtrack(first+1)
                arr[first],arr[i]=arr[i],arr[first]
        # 先排序
        #arr.sort()
        backtrack(0)
        #res.sort()
        return res 
    def Permutation2(self , str: str) -> List[str]:
        # write code here
        #字符串转为数组
        n=len(str)
        arr=list(str)
        res=[]
        def backtrack(arr,start=0):
            if n-1 ==start:
                st="".join(arr)
                if st not in res:
                    res.append(st)
                return 
            st=set()
            for i in range(start,n):
                if arr[i] in st: #跳过已经交换的字符
                    continue
                st.add(arr[i])
                 
                arr[start],arr[i]=arr[i],arr[start]
                
                backtrack(arr,start+1)
                arr[start],arr[i]=arr[i],arr[start]
        # 先排序
        arr.sort()
        backtrack(arr)
        #res.sort()
        return res 
    def Permutation(self , str: str) -> List[str]:
        # write code here
        #字符串转为数组
        n=len(str)
        arr=list(str)
        res=[]
        isvisited=[False]*n
        def backtrack (arr:list,path:list,isvisited:list):
            if len(arr)==len(path):
                st="".join(path.copy())
                res.append(st)
                return 
            for i in range(n):
                if i>0 and arr[i]==arr[i-1] and not isvisited[i-1]: #跳过相同字符
                    continue
                if not isvisited[i]: # 未访问过
                    path.append(arr[i])
                    isvisited[i]=True
                    backtrack(arr,path,isvisited)
                    path.pop()
                    isvisited[i]=False
                
        # 先排序
        arr.sort()
        backtrack(arr,[],isvisited)
        #res.sort()
        return res