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 <= 100s and goal consist of lowercase English letters.
Solutions
Solution 1
PythonJavaC++GoTypeScriptRustPHP
class Solution:
def rotateString(self, s: str, goal: str) -> bool:
return len(s) == len(goal) and goal in s + s(code-box)
class Solution {
public boolean rotateString(String s, String goal) {
return s.length() == goal.length() && (s + s).contains(goal);
}
}(code-box)
class Solution {
public:
bool rotateString(string s, string goal) {
return s.size() == goal.size() && strstr((s + s).data(), goal.data());
}
};(code-box)
func rotateString(s string, goal string) bool {
return len(s) == len(goal) && strings.Contains(s+s, goal)
}(code-box)
function rotateString(s: string, goal: string): boolean {
return s.length === goal.length && (goal + goal).includes(s);
}(code-box)
impl Solution {
pub fn rotate_string(s: String, goal: String) -> bool {
s.len() == goal.len() && (s.clone() + &s).contains(&goal)
}
}(code-box)
class Solution {
/**
* @param String $s
* @param String $goal
* @return Boolean
*/
function rotateString($s, $goal) {
return strlen($goal) === strlen($s) && strpos($s . $s, $goal) !== false;
}
}(code-box)
