484. Find Permutation (Medium)
By now, you are given a secret signature consisting of character 'D' and 'I'. 'D' represents a decreasing relationship between two numbers, 'I' represents an increasing relationship between two numbers. And our secret signature was constructed by a special integer array, which contains uniquely all the different number from 1 to n (n is the length of the secret signature plus 1). For example, the secret signature "DI" can be constructed by array [2,1,3] or [3,1,2], but won't be constructed by array [3,2,4] or [2,1,3,4], which are both illegal constructing special string that can't represent the "DI" secret signature.
On the other hand, now your job is to find the lexicographically smallest permutation of [1, 2, ... n] could refer to the given secret signature in the input.
Example 1:
Input:
"I"
Output:
[1,2]
Explanation:
[1,2] is the only legal initial spectial string can construct secret signature "I", where the number 1 and 2 construct an increasing relationship.
Example 2:
Input:
"DI"
Output:
[2,1,3]
Explanation:
Both [2,1,3] and [3,1,2] can construct the secret signature "DI",
but since we want to find the one with the smallest lexicographical permutation, you need to output [2,1,3]
Note:
The input string will only contain the character 'D' and 'I'.
The length of input string is a positive integer and will not exceed 10,000
Solution: Greedy
I have used a greedy algorithm:
- Loop on the input and insert a decreasing numbers when see a 'I'
- Insert a decreasing numbers to complete the result.
Simple example:
Input: "DIDDID"
0 []
1 [2, 1]
2 [2, 1]
3 [2, 1]
4 [2, 1, 5, 4, 3]
5 [2, 1, 5, 4, 3]
[2, 1, 5, 4, 3, 7, 6]
Then, output is [2, 1, 5, 4, 3, 7, 6]
vector<int> findPermutation(string s) {
vector<int> ret;
for (int i = 0; i <= s.size(); ++i)
if (i == s.size() || s[i] == 'I')
for (int j = i + 1, lenTmp = ret.size(); j > lenTmp; --j)
ret.push_back(j);
return ret;
}