[Leetcode]942. DI String Match
- Easy
- Related Topic: Math
Problem
Given a string S that only contains “I” (increase) or “D” (decrease), let N = S.length.
Return any permutation A of [0, 1, …, N] such that for all i = 0, …, N-1:
If S[i] == “I”, then A[i] < A[i+1]
If S[i] == “D”, then A[i] > A[i+1]
Example 1:
Input: “IDID”
Output: [0,4,1,3,2]
Example 2:
Input: “III”
Output: [0,1,2,3]
Example 3:
Input: “DDI”
Output: [3,2,0,1]
Note:
1 <= S.length <= 10000
S only contains characters “I” or “D”.
- O(N)
class Solution:
def diStringMatch(self, S):
"""
:type S: str
:rtype: List[int]
"""
res = []
l = len(S)
h = 0
t = l
for g in S:
if g == "I":
res.append(h)
h += 1
if g == "D":
res.append(t)
t -= 1
res.append(h)
return res