942. DI String Match
Easy
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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"
.
解法:这个算今天做到最简单的题了。
class Solution {
public:
vector<int> diStringMatch(string S) {
vector<int> ans;
int N=S.length();
int hi=N;
int lo=0;
for(int j=0;j<=N;j++)
{
if(S[j]=='I')ans.push_back(lo++);
else ans.push_back(hi--);
}
return ans;
}
};