Given an input string, reverse the string word by word.
For example,
Given s = "the sky is blue",
return "blue is sky the".
并remove extra spaces
Solution1:按char操作
思路: 整体全部chars reverse + every words reverse
Time Complexity: O(N) Space Complexity: O(N) 算返回的result
Solution2:按word操作,逆序输出
思路:按word操作,逆序输出
Time Complexity: O(N) Space Complexity: O(N) 算返回的result
Solution1 Code:
public 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;
}
}
}
Solution2 Code:
public class Solution2 {
public String reverseWords(String s) {
String[] tmp = s.split("\\s");
StringBuilder sb = new StringBuilder();
for(int i = 1; i <= tmp.length; i++){
if(tmp[tmp.length - i].equals("")){
continue;
}
sb.append(tmp[tmp.length - i]);
sb.append(" ");
}
return sb.toString().trim();
}
}
or
public class Solution2 {
public String reverseWords(String s) {
String[] words = s.trim().split(" +");
Collections.reverse(Arrays.asList(words));
return String.join(" ", words);
}
}