编程之法*反转单词

/*题目描述：翻转句子中单词的顺序，但单词内字符的顺序不变。句子中单词以空格符隔开。为简单起见，标点符号和普通字母一样处理。如："I am a student."翻转成"student. a am I".解题思路：先将整个字符串反转，然后对单个单词进行反转，根据空格来判断一个单词 反转整个串的时间开销为 n/2,空间开销 1. 时间复杂度为O(n)，空间复杂度为O(1). */#include<stdio.h>#include<stdlib.h>#include<string.h>void reverseString(char *s,int from,int to);void reverseWord(char *s);void reverseString(char *s,int _from,int _to) {int from = _from;int to = _to;while ( from < to) {char t = s[from];s[from++] = s[to];s[to--] = t;} } void reverseWord(char *s) {int sigwrdStart = 0;int sigwrdEnd = 0;//反转单个单词 for (int i=0;i < strlen(s)-1;i++) {sigwrdEnd = i;if (s[i] == ' ' && s[i] != '\0' )  //每个单词之间隔一个空格 {reverseString(s,sigwrdStart,sigwrdEnd-1);//反转单词 sigwrdStart = sigwrdEnd + 1; //跳到下一个单词进行反转 } }} int main() {char str[] = "I am a student.";reverseString(str,0,strlen(str)-1);reverseWord(str);printf("%s\n",str);return 0;}