Easy
A permutation perm of n + 1 integers of all the integers in the range [0, n] can be represented as a string s of length n where:
s[i] == 'I'ifperm[i] < perm[i + 1], ands[i] == 'D'ifperm[i] > perm[i + 1].
Given a string s, reconstruct the permutation perm and return it. If there are multiple valid permutations perm, return any of them.
Example 1:
Input: s = "IDID" Output: [0,4,1,3,2]
Example 2:
Input: s = "III" Output: [0,1,2,3]
Example 3:
Input: s = "DDI" Output: [3,2,0,1]
Constraints:
1 <= s.length <= 105s[i]is either'I'or'D'.
class Solution:
def diStringMatch(self, s: str) -> List[int]:
l, h = 0, len(s)
res = []
for i, c in enumerate(s):
if c == "I":
res.append(l)
l+=1
else:
res.append(h)
h-=1
return res+[l]