代码随想录第二十一天

字符串-反转字符串

题目链接

class Solution {
public:
    void reverseString(vector<char>& s) {
        int fast = s.size() - 1;
        int slow = 0;
        while(fast > slow){    //fast != slow会出错,当字符串是偶数的时候会出事，fast与slow永远不会相等
            char tmp = s[slow];
            s[slow] = s[fast];
            s[fast] = tmp;
            fast--;
            slow++;
        }
    }
};

//可以直接用swap
class Solution {
public:
    void reverseString(vector<char>& s) {
        for (int i = 0, j = s.size() - 1; i < s.size()/2; i++, j--) {
            swap(s[i],s[j]);
        }
    }
};

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        left, right = 0, len(s) - 1
        
        # 该方法已经不需要判断奇偶数，经测试后时间空间复杂度比用 for i in range(right//2)更低
        # 推荐该写法，更加通俗易懂
        while left < right:
            s[left], s[right] = s[right], s[left]
            left += 1
            right -= 1