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:

1. Loop on the input and insert a decreasing numbers when see a 'I'
2. 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;
}