






class Solution {
public String reverseWords(String s) {
StringBuilder sb = removeSpace(s);
reverseString(sb, 0, sb.length() - 1);
reverseEachWord(sb);
return sb.toString();
}
private StringBuilder removeSpace(String s) {
int start = 0;
int end = s.length() - 1;
while (s.charAt(start) == ' ') start++;
while (s.charAt(end) == ' ') end--;
StringBuilder sb = new StringBuilder();
while (start <= end) {
char c = s.charAt(start);
if (c != ' ' || sb.charAt(sb.length() - 1) != ' ') {
sb.append(c);
}
start++;
}
return sb;
}
public void reverseString(StringBuilder sb, int start, int end) {
while (start < end) {
char temp = sb.charAt(start);
sb.setCharAt(start, sb.charAt(end));
sb.setCharAt(end, temp);
start++;
end--;
}
}
private void reverseEachWord(StringBuilder sb) {
int start = 0;
int end = 1;
int n = sb.length();
while (start < n) {
while (end < n && sb.charAt(end) != ' ') {
end++;
}
reverseString(sb, start, end - 1);
start = end + 1;
end = start + 1;
}
}
}

class Solution {
public String reverseLeftWords(String s, int n) {
StringBuilder sb=new StringBuilder();
int start=0, end=s.length();
sb.append(s.substring(n,end));
sb.append(s.substring(start,n));
return sb.toString();
}
}

class Solution {
public int strStr(String haystack, String needle) {
if(needle.length()==0){
return 0
}
int [] res =new int[needle.length()]
getPrefix(res, needle)
int j=0
for(int i=0
while(haystack.charAt(i)!=needle.charAt(j)&&j>0){
j=res[j-1]
}
if(haystack.charAt(i)==needle.charAt(j)){
j++
}
if(j==needle.length()){
return i-needle.length()+1
}
}
return -1
}
public int [] getPrefix(int [] res,String needle){
//int []res=new int[needle.length()]
res[0]=0
int j=0
for(int i=1
while(needle.charAt(i)!=needle.charAt(j)&&j>0){
j=res[j-1]
}
if(needle.charAt(i)==needle.charAt(j)) j++
res[i]=j
}
return res
}
}

class Solution {
public boolean repeatedSubstringPattern(String s) {
if (s.equals("")) return false
int len = s.length()
// 原串加个空格(哨兵),使下标从1开始,这样j从0开始,也不用初始化了
s = " " + s
char[] chars = s.toCharArray()
int[] next = new int[len + 1]
// 构造 next 数组过程,j从0开始(空格),i从2开始
for (int i = 2, j = 0
// 匹配不成功,j回到前一位置 next 数组所对应的值
while (j > 0 && chars[i] != chars[j + 1]) j = next[j]
// 匹配成功,j往后移
if (chars[i] == chars[j + 1]) j++
// 更新 next 数组的值
next[i] = j
}
// 最后判断是否是重复的子字符串,这里 next[len] 即代表next数组末尾的值
if (next[len] > 0 && len % (len - next[len]) == 0) {
return true
}
return false
}
}