[LeetCOde] Reverse Words in a String III

 

Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.

Example 1:

Input: "Let's take LeetCode contest"
Output: "s'teL ekat edoCteeL tsetnoc"

Note: In the string, each word is separated by single space and there will not be any extra space in the string.

题目要求把一个含有空格符的字符串中的每个单词反转，就需要对整个字符串根据空格符进行分割。对每个分割的单词进行反转，用tmp保存每个单词的首位索引，需要考虑的是最后一个单词后面没有空格，这时就对其进行单独判断后再进行反转。

class Solution {
public:
    string reverseWords(string s) {
        int count = 0, tmp = 0;
        for (int i = 0; i != s.size(); i++) {
            if (i == s.size() - 1) {
                int left = tmp, right = i;
                while (left <= right)
                    swap(s[left++], s[right--]);
            }
            else if (s[i] != ' ')
                count = i;
            else {
                int left = tmp, right = count;
                while (left <= right)
                    swap(s[left++], s[right--]);
                tmp = count + 2;
                count += 1;
            }
        }
        return s;
    }
};
// 19 ms

巧妙的写法：通过迭代器用i, j记录每个单词首末两位的偏移进行反转。

class Solution {
public:
    string reverseWords(string s) {
        for (int i = 0; i != s.size(); i++) {
            if (s[i] != ' ') {
                int j = i;
                for (; j != s.size() && s[j] != ' '; j++) {}
                reverse(s.begin() + i, s.begin() + j);
                i = j - 1;
            }
        }
        return s;
    }
};
// 23 ms

python写法：'sep'.join(seq): 返回一个以分隔符sep连接各个元素后生成的字符串。 x[::-1]: 对每个字符串进行反转。 split(): 对整个字符串根据空字符进行分割。

class Solution(object):
    def reverseWords(self, s):
        """
        :type s: str
        :rtype: str
        """
        return ' '.join(x[::-1] for x in s.split())
# 82 ms