Question

796. 旋转字符串

English Version

题目描述

给定两个字符串, s 和 goal。如果在若干次旋转操作之后，s 能变成 goal ，那么返回 true 。

s 的 旋转操作 就是将 s 最左边的字符移动到最右边。 

• 例如, 若 s = 'abcde'，在旋转一次之后结果就是'bcdea' 。

 

示例 1:

输入: s = "abcde", goal = "cdeab"
输出: true

示例 2:

输入: s = "abcde", goal = "abced"
输出: false

 

提示:

• 1 <= s.length, goal.length <= 100
• s 和 goal 由小写英文字母组成

解法

方法一

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

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());
  }
};

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);
}

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

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