2020-4-16 第一次 難度：簡單

核心要求：不使用額外的空間，原地修改數(shù)組，O(1)的空間復(fù)雜度

c++代碼

法1:
#include<string>
#include<algorithm>
class Solution {
public:
    void reverseString(vector<char>& s) {
        reverse(s.begin(),s.end());
    }
};


c++法2:
class Solution {
public:
    void reverseString(vector<char>& s) {
        int len = s.size();
        for(int i=0;i<len/2;i++)
        {
            if(s[i]==s[len-1-i]) continue;
            char temp =s[i];
            s[i]=s[len-1-i];
            s[len-1-i]=temp;
        }
    }
};

Python代碼

雙指針法：

    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        i,j = 0,len(s)-1
        while i<j:
            s[i],s[j]=s[j],s[i]
            i+=1
            j-=1

一步法：

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        s[:] = s[::-1]

遞歸+雙指針法：

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        # double pointer and recursion
        def recur_(s, i,j):
            if i>=j:
                return 
            else:
                s[i],s[j] = s[j],s[i]
                recur_(s,i+1,j-1)
        recur_(s,0,len(s)-1)

python代碼參考：反轉(zhuǎn)字符串