找DI序列对应的最小排列

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
vector<int> findPermutation(string s) {
    // 将连续D对应的子段翻转
    const int N = s.size();
    vector<int> ans;
    for (int i = 0; i <= N; i++)
        ans.push_back(i + 1);

    for (int i = 0; i < N; i++) {
        if (s[i] == 'D') {
            int start = i;
            while (i < N && s[i] == 'D')
                i++;
            reverse(ans, start, i);
        }
    }
    return ans;
}

void reverse(vector<int> &v, int left, int right) {
    while (left < right) {
        swap(v[left++], v[right--]);
    }
}

找DI序列对应的任意排列

vector<int> diStringMatch(string S) {
    // 遇'I'输出最小值lo（下一值肯定更大），
    // 遇'D'输出最大值hi（下一值肯定更小）。
    vector<int> ans;
    int lo = 0, hi = S.size();
    for (char c : S) {
        ans.push_back(c == 'I' ? lo++ : hi--);
    }
    ans.push_back(lo);
    return ans;
}

联系：符合DI序列的排列数 - 动态规划