描述
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".
解析
根据题意，找出规律，如果是 "I" 那就是找一个可选值的最小值，如果是 "D" 那就是找一个可选值的最大值，直到找完所有的值。时间复杂度为 O(N)，空间复杂度为 O(N)。
解答
class Solution(object):
def diStringMatch(self, S):
"""
:type S: str
:rtype: List[int]
"""
result = []
arr = [i for i in range(len(S)+1)]
i = 0
j = len(arr)-1
for c in S:
if c=='I':
result.append(arr[i])
i+=1
else:
result.append(arr[j])
j-=1
result.append(arr[i])
return result
运行结果
Runtime: 60 ms, faster than 46.12% of Python online submissions for DI String Match.
Memory Usage: 13.1 MB, less than 44.62% of Python online submissions for DI String Match.
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