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.
把一个字符串中的单词逆序，单词字符顺序不变。
解法1: New array, 新建一个数组，把字符串以空格拆分成单词存到数组，在把单词逆序拷贝进新数组。
解法2: One place，不能新建数组，在原数组的基础上换位。把字符串的所以字符逆序，然后在把每个单词的字符逆序。
Java: New array, two pass
public String reverseWords(String s) {
String[] words = s.trim().split("\\s+");
if(words.length == 0) {
return "";
}
StringBuilder sb = new StringBuilder(words[words.length-1]);
for(int i=words.length-2; i >=0; i--) {
sb.append(" "+words[i]);
}
return sb.toString();
}
Java: New array, one pass
public String reverseWords(String s) {
StringBuilder sb = new StringBuilder();
int end = s.length();
int i = end-1;
while(i>=0) {
if(s.charAt(i) == ' ') {
if(i < end-1) {
sb.append(s.substring(i+1, end)).append(" ");
}
end = i;
}
i--;
}
sb.append(s.substring(i+1, end));
return sb.toString().trim();
}
Java: New array, one pass
public String reverseWords(String s) {
StringBuilder sb = new StringBuilder();
int last = s.length();
for(int i=s.length()-1; i>=-1; i--) {
if(i==-1 || s.charAt(i)==' ') {
String word = s.substring(i+1, last);
if(!word.isEmpty()) {
if(sb.length() != 0) sb.append(' ');
sb.append(word);
}
last = i;
}
}
return sb.toString();
}
Java：One place
public String reverseWords(String s) {
if(s == null || s.isEmpty()) return s;
char[] data = s.toChartArray();
int n = data.length;
reverse(data, 0, n-1);
int last = -1;
for(int i=0; i<=n; i++) {
if(i == n || data[i] == ' ') {
if(i-last>1) reverse(data, last+1, i-1);
last = i;
}
}
return new String(data);
}
private void reverse(char[] data, int start, int end) {
while(start < end) {
char tmp = data[start];
data[start++] = data[end];
data[end--] = tmp;
}
}
Python: New array
class Solution:
# @param s, a string
# @return a string
def reverseWords(self, s):
return ' '.join(reversed(s.split()))
C++:
class Solution {
public:
void reverseWords(string &s) {
int storeIndex = 0, n = s.size();
reverse(s.begin(), s.end());
for (int i = 0; i < n; ++i) {
if (s[i] != ' ') {
if (storeIndex != 0) s[storeIndex++] = ' ';
int j = i;
while (j < n && s[j] != ' ') s[storeIndex++] = s[j++];
reverse(s.begin() + storeIndex - (j - i), s.begin() + storeIndex);
i = j;
}
}
s.resize(storeIndex);
}
};
类似题目：
[LeetCode] 186. Reverse Words in a String II 翻转字符串中的单词 II
[LeetCode] 557. Reverse Words in a String III 翻转字符串中的单词 III
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