leetcode-剑指58-OK

language: C

address

char* reverseWords(char* s){
    int i,p,j;


    // 先算长度
    int getlen(char *st){
        int count = 0;
        while(st[count] != '')
            count++;
        return count;
    }
    int len = getlen(s);
    if(len ==0)
        return "";

    // 翻转的函数，
    void inverse(char *st, int start, int end){
        char temp;
        while(start<end){
            temp = st[start];
            st[start] = st[end];
            st[end] = temp;
            start++;
            end--;
        }
    }


    // 为  去除多余的空格  做的准备
    p = 0; //从此开始p 用来记录空格的数量，最后去尾end = len-1-p
    void deleteonechar(char *st, int position,int end){
        for(;position<end;position++){
            st[position] = st[position+1];
        }
        p++;
    }



    // 去头空格
    i = 0;
    while((i<len-p)&&(s[i] ==' '))
        i++;
    for(j=0;j<i;j++)
        deleteonechar(s,0,len-1-p);


    // 全翻转一次
    inverse(s,0,len-1-p);


    // 把每个单词翻转一次,
    int m=0;
    bool i_filled =false;
    while(m<len-p){
        if(i_filled){   // 找j
            if(s[m]==' '){
                inverse(s,i,m-1);
                i_filled = false;
            }
        }else{
            if(s[m] !=' '){
                i = m;
                i_filled = true;
            }
        }
        m++;
    }
    inverse(s,i,m-1);


    i = 0;
    // 先找不为空格的第一个字符
    while((i<len-p)&&(s[i] ==' '))
        i++;
    for(j=0;j<i;j++)
        deleteonechar(s,0,len-1-p);

    // 接下来去除所有单词中间的多余的空格
    i = 0;
    while(i<len-p){
        if((i-1>=0)&&(s[i-1] == ' ')&&(s[i] == ' ')){
            deleteonechar(s,i,len-1-p);
            i--;
        }
        i++;
    }

    if(p!=0)
        s[len-p] = '';
    return s;
}