QUESTION:
Given a 0-indexed string word and a character ch, reverse the segment of word that starts at index 0 and ends at the index of the first occurrence of ch (inclusive). If the character ch does not exist in word, do nothing.
For example, if word = “abcdefd” and ch = “d”, then you should reverse the segment that starts at 0 and ends at 3 (inclusive). The resulting string will be “dcbaefd”. Return the resulting string.
Example 1:
Input: word = "abcdefd", ch = "d"
Output: "dcbaefd"
Explanation: The first occurrence of "d" is at index 3.
Reverse the part of word from 0 to 3 (inclusive), the resulting string is "dcbaefd".
Example 2:
Input: word = "xyxzxe", ch = "z"
Output: "zxyxxe"
Explanation: The first and only occurrence of "z" is at index 3.
Reverse the part of word from 0 to 3 (inclusive), the resulting string is "zxyxxe".
Example 3:
Input: word = "abcd", ch = "z"
Output: "abcd"
Explanation: "z" does not exist in word.
You should not do any reverse operation, the resulting string is "abcd".
Constraints:
1 <= word.length <= 250
word consists of lowercase English letters.
ch is a lowercase English letter.
EXPLANATION:
easy的题目 一个for循环, 在ch之前的就在前面添加, ch之后的就加载后面即可. 不要忘记刚开始判断是否包含.
SOLUTION:
class Solution {
func reversePrefix(_ word: String, _ ch: Character) -> String {
if !word.contains(ch) {
return word
}
var flag:Bool = true
var result:String = ""
for w in word {
if flag {
result = String(w) + result
} else {
result = result + String(w)
}
if w == ch && flag {
flag = false
}
}
return result
}
}