Merge pull request #2638 from robotsouper/master

更新了Leetcode 131 java部分的写法
This commit is contained in:
程序员Carl
2024-07-23 10:33:14 +08:00
committed by GitHub

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@ -310,39 +310,35 @@ public:
### Java
```Java
class Solution {
List<List<String>> lists = new ArrayList<>();
Deque<String> deque = new LinkedList<>();
//保持前几题一贯的格式, initialization
List<List<String>> res = new ArrayList<>();
List<String> cur = new ArrayList<>();
public List<List<String>> partition(String s) {
backTracking(s, 0);
return lists;
backtracking(s, 0, new StringBuilder());
return res;
}
private void backTracking(String s, int startIndex) {
//如果起始位置大于s的大小说明找到了一组分割方案
if (startIndex >= s.length()) {
lists.add(new ArrayList(deque));
private void backtracking(String s, int start, StringBuilder sb){
//因为是起始位置一个一个加的所以结束时start一定等于s.length,因为进入backtracking时一定末尾也是回文所以cur是满足条件的
if (start == s.length()){
//注意创建一个新的copy
res.add(new ArrayList<>(cur));
return;
}
for (int i = startIndex; i < s.length(); i++) {
//如果是回文子串,则记录
if (isPalindrome(s, startIndex, i)) {
String str = s.substring(startIndex, i + 1);
deque.addLast(str);
} else {
continue;
//像前两题一样从前往后搜索如果发现回文进入backtracking,起始位置后移一位循环结束照例移除cur的末位
for (int i = start; i < s.length(); i++){
sb.append(s.charAt(i));
if (check(sb)){
cur.add(sb.toString());
backtracking(s, i + 1, new StringBuilder());
cur.remove(cur.size() -1 );
}
//起始位置后移,保证不重复
backTracking(s, i + 1);
deque.removeLast();
}
}
//判断是否是回文串
private boolean isPalindrome(String s, int startIndex, int end) {
for (int i = startIndex, j = end; i < j; i++, j--) {
if (s.charAt(i) != s.charAt(j)) {
return false;
}
//helper method, 检查是否是回文
private boolean check(StringBuilder sb){
for (int i = 0; i < sb.length()/ 2; i++){
if (sb.charAt(i) != sb.charAt(sb.length() - 1 - i)){return false;}
}
return true;
}