[LeetCode186]Reverse Words in a String II

Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters.The input string does not contain leading or trailing spaces and the words are always separated by a single space.For example,Given s = "the sky is blue",return "blue is sky the".Could you do it in-place without allocating extra space?Related problem: Rotate ArrayHide Company Tags AmazonHide Tags StringHide Similar Problems (M) Reverse Words in a String (E) Rotate Array

这题好像在career cup上见过。一下子就想到： reverse every word first then reverse the whole sentence.

class Solution {public:    void reverseWords(string &s) {        int start = 0;        reverse(s.begin(), s.end());        for (int i = s.find_first_of(" "); i<s.size(); i = s.find_first_of(" ", i+1)) {            reverse(s.begin()+start, s.begin()+i);            start = i + 1;        }        reverse(s.begin()+start, s.end());    }};

另外一种。

class Solution {public:    void reverseWords(string &s) {        for(int i = 0, j = 0; i<s.size(); i = j+1){            j = i;            while(j<s.size() && !isblank(s[j])) ++j;            reverse(s.begin()+i, s.begin()+j);        }        reverse(s.begin(), s.end());    }};

isblank(): Check if character is blank.

update: 这道题比之前那道(LeetCode151)简单很多，因为之前的会可能有trailing or leading spaces 并且word 之间的space并不是只有一个的。

贴个code：

class Solution {public:    void reverseWords(string &s) {        if(s.empty()) return;        int head = 0, tail = s.size()-1, start = 0;        while(tail >=0 && isblank(s[tail])) --tail;        while(head < s.size() && isblank(s[head])) ++head;        s = s.substr(head, tail-head+1);        for(int i = s.find_first_of(" "); i<s.size(); i = s.find_first_of(" ", i+1)){            while(i<s.size() && isblank(s[i+1])) s.erase(i,1);//delete duplicate " "!            reverse(s.begin()+start, s.begin()+i);            start = i+1;        }        reverse(s.begin()+start, s.end());//reverse the last word first.        reverse(s.begin(), s.end());    }};