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

这道题给了我们一个由D和I两个字符组成的字符串，分别表示对应位置的升序和降序，要我们根据这个字符串生成对应的数字字符串。由于受名字中的permutation的影响，感觉做法应该是找出所有的全排列然后逐个数字验证，这种方法十有八九无法通过OJ。其实这题用贪婪算法最为简单，我们来看一个例子：

D D I I D I

1 2 3 4 5 6 7

3 2 1 4 6 5 7

我们不难看出，只有D对应的位置附近的数字才需要变换，而且变换方法就是倒置一下字符串，我们要做的就是通过D的位置来确定需要倒置的子字符串的起始位置和长度即可。通过观察，我们需要记录D的起始位置i，还有D的连续个数k，那么我们只需要在数组中倒置[i, i+k]之间的数字即可，根据上述思路可以写出代码如下：

解法一：

class Solution {
public:
    vector<int> findPermutation(string s) {
        int n = s.size(), cnt = 0;
        vector<int> res(n + 1);
        for (int i = 0; i < n + 1; ++i) res[i] = i + 1;
        for (int i = 0; i < n; ++i) {
            if (s[i] == 'D') {
                int j = i;
                while (s[i] == 'D' && i < n) ++i;
                reverse(res.begin() + j, res.begin() + i + 1);
                --i;
            } else {
                cnt = 0;
            }
        }
        return res;
    }
};

下面这种方法没有用到数组倒置，而是根据情况来往结果res中加入正确顺序的数字，我们遍历s字符串，遇到D直接跳过，遇到I进行处理，我们每次先记录下结果res的长度size，然后从i+1的位置开始往size遍历，将数字加入结果res中即可，参见代码如下：

解法二：

class Solution {
public:
    vector<int> findPermutation(string s) {
        vector<int> res;
        for (int i = 0; i < s.size() + 1; ++i) {
            if (i == s.size() || s[i] == 'I') {
                int size = res.size();
                for (int j = i + 1; j > size; --j) {
                    res.push_back(j);
                }
            }
        }
        return res;
    }
};

类似题目：

Palindrome Permutation II

Palindrome Permutation

Permutation Sequence

Permutations II

Permutations

Next Permutation

参考资料：

https://discuss.leetcode.com/topic/76230/c-simple-solution-in-72ms-and-9-lines/2

https://discuss.leetcode.com/topic/76213/greedy-o-n-java-solution-with-explanation

https://discuss.leetcode.com/topic/76221/java-o-n-clean-solution-easy-to-understand/2