# 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

class Solution(object):
    def findPermutation(self, s):
        """
        :type s: str
        :rtype: List[int]
        """
        # 利用栈 贪心寻找倒置起止点
        res = []
        stack = []
        
        for i in range(1, len(s)+1):
            if s[i-1] == 'I':
                stack.append(i)
                while stack:
                    res.append(stack.pop())
            else:
                stack.append(i)

        stack.append(len(s)+1)
        while stack:
            res.append(stack.pop())
        return res