leetcode - Reverse Words in a String

leetcode - Reverse Words in a String

 

Given an input string, reverse the string word by word.

For example,
Given s = "the sky is blue",
return "blue is sky the".

Update (2015-02-12):
For C programmers: Try to solve it in-place in O(1) space.

click to show clarification.

Clarification:

 

• What constitutes a word?
  A sequence of non-space characters constitutes a word.
• Could the input string contain leading or trailing spaces?
  Yes. However, your reversed string should not contain leading or trailing spaces.
• How about multiple spaces between two words?
  Reduce them to a single space in the reversed string.

class Solution {
public:
    void reverseWords(string &s) {
        reverse(s.begin(), s.end()); //reverse the whole string.
        string::iterator it = s.begin();
        while(it!=s.end()){
            while(it != s.end() && *it==' '){
                if(*(it+1)==' '){ it = s.erase(it);}
               //if the next one is space, erase the current one.
               // erase will return the next iterator.
                else{it++;} // else it++
            }
            string::iterator beg = it;
            while(it != s.end() && *it!=' '){
                it++;
            }
            string::iterator end = it;
            reverse(beg,end); // reverse each words.
        }
        if(*(s.begin()) == ' ') s.erase(s.begin()); // remove the first space
        if(*(s.end()-1) == ' ') s.erase(s.end()-1); // remove the last space.
    }
};

剑指offer面试题42.

先将整个字符串转置，然后对其中每个词转置。

注意使用erase用法，输入待删除的迭代器，返回的是后面第一个有效的迭代器，之前的后面的迭代器都失效。

注意reverse的用法，输入的是迭代器区间，包括第一个，不包括最后一个（前闭后开区间）。

注意处理多个空格的情况，遍历完后保证只有一个空格，然后清除头尾剩余的空格（如果有的话）。

整个算法in-place.