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描述 ：给定一个字符串 s s s i n d i c e s indices in d i ces i n d i c e s [ i ] indices[i] in d i ces [ i ] s [ i ] s[i] s [ i ]
要求 ：返回重新排列后的字符串。
说明 ：
1 ≤ s . l e n g t h ≤ 100 1 \le s.length \le 100 1 ≤ s . l e n g t h ≤ 100 i n d i c e s indices in d i ces 0 ∼ n − 1 0 \sim n-1 0 ∼ n − 1 示例 ：
输入：s = "codeleet" , indices = [ 4 , 5 , 6 , 7 , 0 , 2 , 1 , 3 ] 输出： "leetcode" 解释：如图所示， "codeleet" 重新排列后变为 "leetcode" 。 输入：s = "abc" , indices = [ 0 , 1 , 2 ] 输出： "abc" 解释：重新排列后，每个字符都还留在原来的位置上。 创建一个长度为 n n n r e s res res i i i s [ i ] s[i] s [ i ] r e s [ i n d i c e s [ i ] ] res[indices[i]] res [ in d i ces [ i ]]
第 1 步 ：r e s = [ ′ ′ ] × n res = [''] \times n res = [ ′′ ] × n
第 2 步 ：遍历 i i i r e s [ i n d i c e s [ i ] ] = s [ i ] res[indices[i]] = s[i] res [ in d i ces [ i ]] = s [ i ]
第 3 步 ：返回 ''.join(res)。
以 s = " c o d e l e e t " , i n d i c e s = [ 4 , 5 , 6 , 7 , 0 , 2 , 1 , 3 ] s = "codeleet", indices = [4,5,6,7,0,2,1,3] s = " co d e l ee t " , in d i ces = [ 4 , 5 , 6 , 7 , 0 , 2 , 1 , 3 ]
s [ 0 ] = ′ c ′ → r e s [ 4 ] = ′ c ′ s[0]='c' \to res[4]='c' s [ 0 ] = ′ c ′ → res [ 4 ] = ′ c ′ s [ 1 ] = ′ o ′ → r e s [ 5 ] = ′ o ′ s[1]='o' \to res[5]='o' s [ 1 ] = ′ o ′ → res [ 5 ] = ′ o ′ s [ 2 ] = ′ d ′ → r e s [ 6 ] = ′ d ′ s[2]='d' \to res[6]='d' s [ 2 ] = ′ d ′ → res [ 6 ] = ′ d ′ s [ 3 ] = ′ e ′ → r e s [ 7 ] = ′ e ′ s[3]='e' \to res[7]='e' s [ 3 ] = ′ e ′ → res [ 7 ] = ′ e ′ s [ 4 ] = ′ l ′ → r e s [ 0 ] = ′ l ′ s[4]='l' \to res[0]='l' s [ 4 ] = ′ l ′ → res [ 0 ] = ′ l ′ s [ 5 ] = ′ e ′ → r e s [ 2 ] = ′ e ′ s[5]='e' \to res[2]='e' s [ 5 ] = ′ e ′ → res [ 2 ] = ′ e ′ s [ 6 ] = ′ e ′ → r e s [ 1 ] = ′ e ′ s[6]='e' \to res[1]='e' s [ 6 ] = ′ e ′ → res [ 1 ] = ′ e ′ s [ 7 ] = ′ t ′ → r e s [ 3 ] = ′ t ′ s[7]='t' \to res[3]='t' s [ 7 ] = ′ t ′ → res [ 3 ] = ′ t ′ 结果："leetcode"。
class Solution : def restoreString ( self , s : str , indices : List[ int ] ) -> str : n = len (s) res = [ '' ] * n for i, idx in enumerate (indices): res[idx] = s[i] return '' . join (res) 时间复杂度 ：O ( n ) O(n) O ( n ) 空间复杂度 ：O ( n ) O(n) O ( n ) 
