在原字符串中把字符串尾部的m个字符移动到字符串的头部。
class Solution {
public void reverse(char[] ch, int l ,int r) {
for ( ; l<r; l++,r--){
char t = ch[l];
ch[l] = ch[r];
ch[r] = t;
}
}
public void fun(String str, int n, int m){
if(str == null || str.length()==0) return;
m %= n;
char[] ch = str.toCharArray();
reverse(ch, 0, m-1);
reverse(ch, m, n-1);
reverse(ch, 0, n-1);
for(int i=0; i<ch.length; i++){
System.out.print(ch[i]);
}
}
public static void main(String[] args) {
String s = "abcdef";
new Solution().fun(s,6,4); //前四个字符移到末尾：efabcd
}
}
链接
25. Reverse Nodes in k-Group
Hard
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Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
Example:
Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5
Note:
Only constant extra memory is allowed.
You may not alter the values in the list's nodes, only nodes itself may be changed.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode pre = dummy;
ListNode end = dummy;
while(end.next != null){
for(int i=0; i<k && end != null; i++) end = end.next;
if(end == null) break;
ListNode start = pre.next;
ListNode next = end.next;
end.next = null;
pre.next = reverse(start);
start.next = next;
pre = start;
end = pre;
}
return dummy.next;
}
public ListNode reverse(ListNode head){
ListNode pre = null;
while(head != null){
ListNode next = head.next;
head.next = pre;
pre = head;
head = next;
}
return pre;
}
}
【151. Reverse Words in a String】
链接
Given an input string, reverse the string word by word.
Example 1:
Input: "the sky is blue"
Output: "blue is sky the"
Example 2:
Input: " hello world! "
Output: "world! hello"
Explanation: Your reversed string should not contain leading or trailing spaces.
Example 3:
Input: "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.
【代码】
class Solution {
public void reverse(char[] ch,int l,int r){
while(l<r){
char t = ch[l];
ch[l++] = ch[r];
ch[r--] = t;
}
}
public void recWord(char[] ch, int n){
int i = 0, j = 0;
while(i < n){
//make i and j equal (j might be ahead as it might have seen a word before)
while(i<j || (i<n && ch[i]==' ')) i++; // skip spaces
//make j and i equal (i might be ahead as it found few spaces)
while(j<i || (j<n && ch[j]!=' ')) j++; // skip non spaces
reverse(ch, i, j-1); // reverse the word
}
}
public String clean(char[] ch, int n){
int i=0, j=0;
while(j<n){
while(j<n && ch[j]==' ') j++; // skip spaces
while(j<n && ch[j]!=' ') ch[i++] = ch[j++]; // keep non spaces
while(j<n && ch[j]==' ') j++;
if(j<n) ch[i++]=' '; // keep only one space
}
return new String(ch).substring(0,i);
}
public String reverseWords(String s) {
if(s==null) return null;
char[] ch = s.toCharArray();
int n = ch.length;
reverse(ch, 0, n-1);
recWord(ch,n);
return clean(ch,n);
}
}