[Leetcode] 484. Find Permutation 解题报告

题目：

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
思路：

很巧妙的一道题目：为了使得permutation最小，我们需要注意D连续出现的次数（不连续的情况就无所谓了，因为有I会调节）。只要在I前面（或者整个字符串的末尾之前）出现了n次D，那么我们就在ret中填入n+1个逆序序列，而这n+1个逆序序列的范围由ret当前的size确定（例如当ret的size为0时，我们就填入n + 1, n, ...1；如果ret的size为2，我们就填入n + 3, n+2,...3）。这样就保证了在D出现的位置上我们都填了尽可能小的数字。整个算法的时间复杂度是O(n)，其中n是字符串s的长度。

代码：

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