484. Find Permutation

By now, you are given asecret signatureconsisting of character 'D' and 'I'. 'D' represents a decreasing relationship between two numbers, 'I' represents an increasing relationship between two numbers. And oursecret signaturewas 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 givensecret signaturein 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

S1: DFS O(n!)

从最小的数字开始尝试直到得到可行解

//超时
class Solution {
public:
    vector<int> findPermutation(string s) {
        int n = s.size();
        vector<int> result(n + 1);
        for (int i = 1; i <= n + 1; ++i) {
            result[0] = i;
            if (dfs(s, 0, 1 << i, result)) return result;
        }
        return vector<int>{};
    }

    bool dfs(string s, int idx, int used, vector<int>& result) {
        if (idx == s.size()) return true;
        int pre = result[idx];
        if (s[idx] == 'D') {
            for (int i = 1; i < pre; ++i) {
                if ((used & (1 << i)) == 0) {
                    result[idx + 1] = i;
                    if(dfs(s, idx + 1, used | (1 << i), result)) return true;
                }
            }
        } else {
            for (int i = pre + 1; i <= s.size() + 1; ++i) {
                if ((used & (1 << i)) == 0) {
                    result[idx + 1] = i;
                    if (dfs(s, idx + 1, used | (1 << i), result)) return true;
                }
            }
        }
        return false;
    }
};

S2: greedy O(n)

构造递增数列，若s全为‘I'则该数列即所求，若有D, 将D多对应得数列+1 反转即为所求

12345 : IDDI -> 14325：IDDI

123: DI -> 213: DI

    vector<int> findPermutation(string s) {
        int n = s.size();
        vector<int> result(n + 1);
        for (int i = 1; i <= n + 1; ++i) {  //构造递增数列
            result[i - 1] = i;
        }
        for (int i = 0; i < n; ++i) {
            int j = i;
            while (j < n && s[j] == 'D') j++;
            if (j > i) {
                reverse(result.begin() + i, result.begin() + j + 1);
                i = j - 1;
            }
        }
        return result;
    }