Description
Given an input string, reverse the string word by word.
Example
Example 1:1
2Input: "the sky is blue"
Output: "blue is sky the"
Example 2:1
2
3Input: " hello world! "
Output: "world! hello"
Explanation: Your reversed string should not contain leading or trailing spaces.
Example 3:1
2
3Input: "a good example"
Output: "example good a"
Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.
Note:
- A word is defined as a sequence of non-space characters.
- Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
- You need to reduce multiple spaces between two words to a single space in the reversed string.
Follow up:
For C programmers, try to solve it in-place in O(1) extra space.
Solution
1 | class Solution { |
Method 2: Don’t use split/trim1
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49public class Solution {
public String reverseWords(String s) {
if (s == null) return null;
char[] a = s.toCharArray();
int n = a.length;
// step 1. reverse the whole string
reverse(a, 0, n - 1);
// step 2. reverse each word
reverseWords(a, n);
// step 3. clean up spaces
return cleanSpaces(a, n);
}
void reverseWords(char[] a, int n) {
int i = 0, j = 0;
while (i < n) {
while (i < j || i < n && a[i] == ' ') i++; // skip spaces
while (j < i || j < n && a[j] != ' ') j++; // skip non spaces
reverse(a, i, j - 1); // reverse the word
}
}
// trim leading, trailing and multiple spaces
String cleanSpaces(char[] a, int n) {
int i = 0, j = 0;
while (j < n) {
while (j < n && a[j] == ' ') j++; // skip spaces
while (j < n && a[j] != ' ') a[i++] = a[j++]; // keep non spaces
while (j < n && a[j] == ' ') j++; // skip spaces
if (j < n) a[i++] = ' '; // keep only one space
}
return new String(a).substring(0, i);
}
// reverse a[] from a[i] to a[j]
private void reverse(char[] a, int i, int j) {
while (i < j) {
char t = a[i];
a[i++] = a[j];
a[j--] = t;
}
}
}