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.
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.
题目解析：1、生成的单词里面没有空格！
2、输入的String，包括头或尾空格？ 是的，反转的时候不能包含这些空格
3、两个单词之间的多个空格？ 只保留一个空格分析：两种方法：
代码1：
String[] parts = s.trim().split("\s+");//去除前后的空格+去除里面的空格，得到一个没有空格的字符串
String out = "";
if (parts.length > 0) { //从后往前输出，加入字符串
for (int i = parts.length - 1; i > 0; i--)
{ out += parts[i] + " "; }
out += parts[0]; }
return out;
代码2：通过反转整个字符串，去除不符合题目的空格；
再反转每一个单词；
package leetcode;
public class ReverseWordsinaString {
// reverses the part of an array and returns the input array for convenience
public static 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 static String reverseWords(String s) {
// reverse the whole string and convert to char array
char[] str = reverse(s.toCharArray(), 0, s.length()-1);
// System.out.println(str.toString());
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
}
}
// System.out.println(s);
// System.out.println(s.length());
reverse(str, start, end-1);
//为什么要加上面那一步呢？？？貌似没有区别
// 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
// System.out.println(s);
// System.out.println(s.length());
return new String(str, 0, end > 0 && str[end-1] == ' ' ? end-1 : end);
}
public static void main(String[] args) {
System.out.println(reverseWords(" he ll o world!! "));
System.out.println(reverseWords(" he ll o world!! ").length());
}
}