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leetcode:反转字符串中的单词III

article 2024/10/12 0:01:13

题目链接

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string reverse(string s1)
{
	string s2;
	string::reverse_iterator rit = s1.rbegin();
	while (rit != s1.rend())
	{
		s2 += *rit;
		rit++;
	}
	return s2;
}
class Solution {
public:
    string reverseWords(string s) {
        string s1;
int i = 0;
int j = 0;
int length = s.length();
for (i = 0; i < length; )//因为不知道具体单词长度，这里不做++处理
{
	while (s[j] != ' ')//找到下一个空格
	{
		j++;
		if (j >= length)
		{
			break;
		}
	}
	int k = j - i;//所需反转的单词长度
	string tmp = s.substr(i, k);
	tmp = reverse(tmp);
	s1 += tmp;

	int m = s.find(' ');
	s1 += ' ';

	i += k+1;
	j = i+1;
	
}
//最后得出的s1末尾会多一个空格，把它截断
s1 = s1.substr(0, s1.length() - 1);
        return s1;
    }
};