【LeetCode】344. Reverse String 解答・解説【Python】

問題
1. 原文
2. 内容
解答
補足・参考・感想

問題

原文

Write a function that reverses a string. The input string is given as an array of characters s.

You must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Input: s = ["h","e","l","l","o"] Output: ["o","l","l","e","h"]

1
2

Input: s = ["h","e","l","l","o"]
Output: ["o","l","l","e","h"]

Example 2:

Input: s = ["H","a","n","n","a","h"] Output: ["h","a","n","n","a","H"]

1
2

Input: s = ["H","a","n","n","a","h"]
Output: ["h","a","n","n","a","H"]

Constraints:

1 <= s.length <= 10⁵

s[i] is a printable ascii character.

内容

文字列を反転させる関数を書いてください。入力される文字列は文字列sの配列で与えられます。

空間計算量O(1)、インプレイス処理で実装してください。

※正しくない可能性があります。

解答

解答1：

class Solution:
    def reverseString(self, s: List[str]) -> None:
        for i in range(len(s)//2):
            s[i], s[-(i+1)] = s[-(i+1)], s[i]

class Solution:

def reverseString(self, s: List[str]) -> None:

for i in range(len(s)//2):

s[i], s[-(i+1)] = s[-(i+1)], s[i]

two pointerの方が速い。

解答2：Python、two pointer

class Solution:
    def reverseString(self, s: List[str]) -> None:
        left = 0
        right = len(s)-1
        while left < right:
            s[left], s[right] = s[right], s[left]
            left += 1
            right -= 1

class Solution:

def reverseString(self, s: List[str]) -> None:

left = 0

right = len(s)-1

while left < right:

s[left], s[right] = s[right], s[left]

left += 1

right -= 1

解答3：Rust、reverse()

impl Solution {
    pub fn reverse_string(s: &mut Vec<char>) {
        s.reverse()       
    }
}

impl Solution {

pub fn reverse_string(s: &mut Vec<char>) {

s.reverse()

}

Pythonでも同じように書ける。

解答4:Rust、iterator

impl Solution {
    pub fn reverse_string(s: &mut Vec<char>) {
        match s.len() {
            0 => (),
            size => {
                (0..(size / 2)).into_iter().for_each(|beg| {
                    let end = size -1 -beg;
                    s.swap(beg, end);
                });
            }
        };
    }
}

impl Solution {

pub fn reverse_string(s: &mut Vec<char>) {

match s.len() {

0 => (),

size => {

(0..(size / 2)).into_iter().for_each(|beg| {

let end = size -1 -beg;

s.swap(beg, end);

});

}

};

}

補足・参考・感想

■補足

■参考

■感想

空間計算量の掴み方がわからない。

前：167. Two Sum II – Input Array Is Sorted

次：19. Remove Nth Node From End of List

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