編寫一個函數,其作用是將輸入的字符串反轉過來。輸入字符串以字符數組 char[]
的形式給出。
不要給另外的數組分配額外的空間,你必須原地修改輸入數組、使用 O(1) 的額外空間解決這一問題。
你可以假設數組中的所有字符都是 ASCII 碼錶中的可打印字符。
示例 1:
輸入:["h","e","l","l","o"]
輸出:["o","l","l","e","h"]
示例 2:
輸入:["H","a","n","n","a","h"]
輸出:["h","a","n","n","a","H"]
解法一:直接利用原生數組的reverse方法
/**
* @param {character[]} s
* @return {void} Do not return anything, modify s in-place instead.
*/
var reverseString = function(s) {
return s.reverse()
};
解法二:遍歷交換首尾相應的位置
/**
* @param {character[]} s
* @return {void} Do not return anything, modify s in-place instead.
*/
var reverseString = function (s) {
for (let index = 0, len = s.length; index < len /2; index++) {
const temp = s[index]
s[index] = s[len - index - 1]
s[len - index - 1] = temp
}
return s
};
解法三:遍歷連接字符串
/**
* @param {character[]} s
* @return {void} Do not return anything, modify s in-place instead.
*/
var reverseString = function (s) {
let len = s.length, str = ''
for (let i = 0; i < len; i++) {
str = s[i] + str
}
return str
};
如有更好的解法,歡迎留言探討。