796. Rotate String

Description

Given two strings s and goal, return true if and only if s can become goal after some number of shifts on s.

A shift on s consists of moving the leftmost character of s to the rightmost position.

• For example, if s = "abcde", then it will be "bcdea" after one shift.

Example 1:

Input: s = "abcde", goal = "cdeab"
Output: true


Example 2:

Input: s = "abcde", goal = "abced"
Output: false


Constraints:

• 1 <= s.length, goal.length <= 100
• s and goal consist of lowercase English letters.

Solutions

• class Solution {
public boolean rotateString(String s, String goal) {
return s.length() == goal.length() && (s + s).contains(goal);
}
}

• class Solution {
public:
bool rotateString(string s, string goal) {
return s.size() == goal.size() && strstr((s + s).data(), goal.data());
}
};

• class Solution:
def rotateString(self, s: str, goal: str) -> bool:
return len(s) == len(goal) and goal in s + s


• func rotateString(s string, goal string) bool {
return len(s) == len(goal) && strings.Contains(s+s, goal)
}

• function rotateString(s: string, goal: string): boolean {
return s.length === goal.length && (goal + goal).includes(s);
}


• class Solution {
/**
* @param String $s * @param String$goal
* @return Boolean
*/
function rotateString($s,$goal) {
return strlen($goal) === strlen($s) && strpos($s .$s, \$goal) !== false;
}
}

• impl Solution {
pub fn rotate_string(s: String, goal: String) -> bool {
s.len() == goal.len() && (s.clone() + &s).contains(&goal)
}
}