484. Find Permutation
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:
Rust Solution
struct Solution;
impl Solution {
fn find_permutation(s: String) -> Vec<i32> {
let n = s.len();
let mut res: Vec<i32> = (1..=(n + 1) as i32).collect();
let s: Vec<char> = s.chars().collect();
let mut l = 0;
while l < n {
if s[l] == 'D' {
let mut r = l;
while r < n && s[r] == 'D' {
r += 1;
}
res[l..=r].reverse();
l = r;
} else {
l += 1;
}
}
res
}
}
#[test]
fn test() {
let s = "I".to_string();
let res = vec![1, 2];
assert_eq!(Solution::find_permutation(s), res);
let s = "DI".to_string();
let res = vec![2, 1, 3];
assert_eq!(Solution::find_permutation(s), res);
}
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