Split palindrome
Problem Description
Given a string s, split s into some substrings so that each substring is a palindrome.
Return all possible splitting schemes for s.
Example:
answer
Backtracking
Backtracking method: an exhaustive method of optimization achieved by depth-first search.
template:
Recursive backtracking:
void backtrack (参数){
if (终止条件) output();
else{
for (int i=f(n,t);i<=g(n,t);i++) {
处理节点;
if (constraint()&&bound()) backtrack();//如果满足约束条件则递归
回溯，撤销对节点的处理;
}
}
}
Iterative backtracking:
void iterativeBacktrack (){
int t=1;//t为层数
while (t>0) {
if (f(n,t)<=g(n,t))
for (int i=f(n,t);i<=g(n,t);i++) {
x[t]=h(i);
if (constraint(t)&&bound(t)) {
if (solution(t)) output(x);
else t++;
}
}
else t--;
}
}
The code of this question:
public class Solution {
public List<List<String>> partition(String s) {
int len = s.length();
List<List<String>> res = new ArrayList<>();
if (len == 0) {
return res;
}
Deque<String> path = new ArrayDeque<>();
backtracking(s, 0, len, path, res);
return res;
}
public void backtracking(String s, int start, int len, Deque<String> path, List<List<String>> res) {
if (start == len) {
//终止条件
res.add(new ArrayList<>(path));
return;
}else{
for (int i = start; i < len; i++) {
if (isPalindrome(s, start, i)) {
//如果是回文串则递归
path.addLast(s.substring(start, i + 1));
backtracking(s, i + 1, len, path, res);
path.removeLast();
}
}
}
}
public boolean isPalindrome(String s, int low, int high) {
while (low < high) {
if (s.charAt(low) != s.charAt(high)) {
return false;
}
low++;
high--;
}
return true;
}
}
Valid letter variants
Problem Description
Given two strings s and t, write a function to determine whether t is an anagram of s.
Description:
- Aliphatic words refer to strings with the same letters but different arrangements.
- You can assume that the string contains only lowercase letters.
Advanced
- What if the input string contains unicode characters? Can you adjust your solution to deal with this situation?
answer
hash table
Drawing solution algorithm: 242. Effective letter dysphorisms
thought:
- First judge whether the two strings are equal in length, and return false directly if they are not equal
- If they are equal, initialize the 26-letter hash table and traverse the strings s and t
- s is responsible for increasing in the corresponding position, t is responsible for decreasing in the corresponding position
- If the value of the hash table is 0, then the two are letter dysphoric words
class Solution {
public boolean isAnagram(String s, String t) {
if(s.length() != t.length())
return false;
int[] alpha = new int[26];
for(int i = 0; i< s.length(); i++) {
alpha[s.charAt(i) - 'a'] ++;
alpha[t.charAt(i) - 'a'] --;
}
for(int i=0;i<26;i++)
if(alpha[i] != 0)
return false;
return true;
}
}
作者：guanpengchn
链接：https://leetcode-cn.com/problems/valid-anagram/solution/hua-jie-suan-fa-242-you-xiao-de-zi-mu-yi-wei-ci-by/
来源：力扣（LeetCode）
Sort
This solution comes from the official solution
Thought: t is an ectopic word of s equivalent to "two strings are equal after sorting." Therefore, we can sort the strings s and t separately and judge whether the sorted strings are equal. In addition, if the lengths of s and tt are different, t must not be an ectopic word of s.
class Solution {
public boolean isAnagram(String s, String t) {
if (s.length() != t.length()) {
return false;
}
char[] str1 = s.toCharArray();
char[] str2 = t.toCharArray();
Arrays.sort(str1);
Arrays.sort(str2);
return Arrays.equals(str1, str2);
}
}