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"
, thenA[i] < A[i+1]
- If
S[i] == "D"
, thenA[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"
.
Approach #1:
class Solution { public: vector<int> diStringMatch(string S) { int len = S.length(); vector<int> ans(len+1, 0); int index = 0; for (int i = 0; i < len; ++i) if (S[i] == 'I') ans[i] = index++; ans[len] = index++; for (int i = len-1; i >= 0; --i) if (S[i] == 'D') ans[i] = index++; return ans; } };