前言:
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不要给另外的数组分配额外的空间,你必须原地修改输入数组、使用 O(1) 的额外空间解决这一问题。
示例 1:
输入:s = ["h","e","l","l","o"]
输出:["o","l","l","e","h"]
示例 2:
输入:s = ["H","a","n","n","a","h"]
输出:["h","a","n","n","a","H"]
提示:
1 <= s.length <= 105
s[i] 都是 ASCII 码表中的可打印字符
来源:力扣(LeetCode)
链接:
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题解
Swift
class Solution {
func reverseString(_ s: inout [Character]) {
let length = s.count
if length < 2 {
return
}
var temp: Character
for index in 0 ..< length / 2 {
temp = s[index]
s[index] = s[length - index - 1]
s[length - index - 1] = temp
}
}
}
var chat: [Character] = ["h", "e", "l", "l", "o"]
print(Solution().reverseString(&chat))
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