【leetcode】Reverse Words in a String(hard)☆

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

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

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

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：翻转的思路是很清楚的，就是卡在空格上了。结果专门先循环一遍来去掉空格。

注意，必须改变指针中所对应的值才能改变字符串。

void reverse(char * s, char * e)

{

    while(s < e)

    {

        char tmp = *s;

        *s = *e;

        *e = tmp;

        s++; e--;

    }

}



void reverseWords(char *s) {

    //先专门处理空格

    char * p = s;

    char * snew = s;

    while(*p != '\0')

    {

        if(*p != ' ') *snew++ = *p++;

        else if(snew == s) p++; //开始处遇到空格

        else if(*(snew - 1) == ' ') p++; //已经有了一个空格

        else *snew++ = *p++;

    }

    *snew = '\0';

    if(*(snew - 1) == ' ') *(snew - 1) = '\0';



    //翻转

    char *ss = s;

    int start = 0, end = 0;

    while(*ss != '\0')

    {

        while(*ss != ' ' && *ss != '\0')

        {

            ss++;  end++;

        }

        reverse(s + start, s + end - 1);

        if(*ss != '\0')

        {

            ss++; end++; start = end;

        }

    }

    reverse(s, s + end - 1);

}

 

看看大神的：不用先去除空格，而是在遍历的过程中用end来更新字符串，去掉空格。

// reverses the part of an array and returns the input array for convenience

public char[] reverse(char[] arr, int i, int j) {

    while (i < j) {

        char tmp = arr[i];

        arr[i++] = arr[j];

        arr[j--] = tmp;

    }

    return arr;

}



public String reverseWords(String s) {

    // reverse the whole string and convert to char array

    char[] str = reverse(s.toCharArray(), 0, s.length()-1);

    int start = 0, end = 0; // start and end positions of a current word

    for (int i = 0; i < str.length; i++) {

        if (str[i] != ' ') { // if the current char is letter 

            str[end++] = str[i]; // just move this letter to the next free pos

        } else if (i > 0 && str[i-1] != ' ') { // if the first space after word

            reverse(str, start, end-1); // reverse the word

            str[end++] = ' '; // and put the space after it

            start = end; // move start position further for the next word

        }

    }

    reverse(str, start, end-1); // reverse the tail word if it's there

    // here's an ugly return just because we need to return Java's String

    // also as there could be spaces at the end of original string 

    // we need to consider redundant space we have put there before

    return new String(str, 0, end > 0 && str[end-1] == ' ' ? end-1 : end);

}