leetcode翻转字符串里的单词

字符串反转技巧

最新推荐文章于 2023-12-17 11:55:38 发布

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本文探讨了一种使用栈来反转字符串中单词顺序的算法。通过实例演示了如何去除多余空格并保持单词间的单个空格，确保输出整洁无冗余。

Example 1:

Input: “the sky is blue”
Output: “blue is sky the”
Example 2:

Input: " hello world! "
Output: “world! hello”
Explanation: Your reversed string should not contain leading or trailing spaces.
Example 3:

Input: “a good example”
Output: “example good a”
Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.

Note:

A word is defined as a sequence of non-space characters.
Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
You need to reduce multiple spaces between two words to a single space in the reversed string.

Follow up:

For C programmers, try to solve it in-place in O(1) extra space.

那种翻转之后再翻转的方法比较主流，这里是用栈实现的一种方法。当然这种方法就借助了其他空间。

class Solution {
public:
    string reverseWords(string s) {
        if(s.size() == 0) return "";
        if (!s.empty())
		{
			s.erase(0, s.find_first_not_of(" "));
			s.erase(s.find_last_not_of(" ") + 1);
		}
        stack<string> s1;
        for(int i = 0; i < s.size(); ++i) {
            while(s[i] == ' ') ++i;
            int j = i; 
            while(s[i] != ' ' && s[i]) ++i;
            s1.push(s.substr(j, i - j));
        }
        string res;
        while(!s1.empty()) {
            res.append(s1.top());
            if(s1.size() != 1)res.append(" ");
            s1.pop();
        }
        return res;
    }
};