[Leetcode] 151. Reverse Words in a String 解题报告

题目：

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.

思路：

刷题刷久了，就忘了面试中关键的一个考察点，就是对一个题目有疑惑的时候，一定要记得和面试官讨论，进行clarification。上面写到的三点clarification都是非常重要的，直接关系到代码该怎么写。

其实算法的思路也算是比较明确的：1）每当检测出来一个单词时，对其进行翻转；2）当所有的单词被检测出来时，对整个字符串再进行一次翻转。由于原字符串中有可能存在多余的空格，所以我们在扫描过程中，需要在字符串内部进行复制，以将多余的空格给覆盖掉。具体见下面的代码片段。该思路的时间复杂度是O(n)，空间复杂度是O(1)。

代码：

class Solution {public:    void reverseWords(string &s) {        int len = s.size();        int left = 0, right = 0, i = 0, flag = false;        while(i < len) {            while(s[i] == ' ')  // trim the leading space                i++;            if(i == len)        // already at the end                break;            if(flag)                s[right++] = ' ';            left = right;            while(i < len && s[i] != ' ')                s[right++] = s[i++];            reverse(s, left, right-1);            flag = true;    // flag is used to judge whether append " "        }        s.resize(right);        reverse(s, 0, right - 1);    }private:    void reverse(string&s, int left, int right) {        while(left < right) {            char ch = s[right];            s[right--] = s[left];            s[left++] = ch;        }    }};