给定只含 "I"
(增大)或 "D"
(减小)的字符串 S
,令 N = S.length
。
返回 [0, 1, ..., N]
的任意排列 A
使得对于所有 i = 0, ..., N-1
,都有:
- 如果
S[i] == "I"
,那么A[i] < A[i+1]
- 如果
S[i] == "D"
,那么A[i] > A[i+1]
示例 1:
输出:"IDID" 输出:[0,4,1,3,2]
示例 2:
输出:"III" 输出:[0,1,2,3]
示例 3:
输出:"DDI" 输出:[3,2,0,1]
提示:
1 <= S.length <= 1000
S
只包含字符"I"
或"D"
。
题目标签:Math
题目链接:LeetCode / LeetCode中国
Language | Runtime | Memory |
---|---|---|
python3 | 92 ms | 14.3 MB |
class Solution:
def diStringMatch(self, S: str) -> List[int]:
A = [x for x in range(len(S) + 1)]
res = []
p, q = 0, len(S)
for i in range(len(S)):
if S[i] == 'I':
res.append(A[p])
p += 1
else:
res.append(A[q])
q -= 1
res.append(A[p])
return res