@@ -314,6 +314,47 @@ class Solution {
314314 return true ;
315315 }
316316}
317+ // 方法一:但使用stringBuilder,故优化时间、空间复杂度,因为向字符串插入字符时无需复制整个字符串,从而减少了操作的时间复杂度,也不用开新空间存subString,从而减少了空间复杂度。
318+ class Solution {
319+ List<String > result = new ArrayList<> ();
320+ public List<String > restoreIpAddresses (String s ) {
321+ StringBuilder sb = new StringBuilder (s);
322+ backTracking(sb, 0 , 0 );
323+ return result;
324+ }
325+ private void backTracking (StringBuilder s , int startIndex , int dotCount ){
326+ if (dotCount == 3 ){
327+ if (isValid(s, startIndex, s. length() - 1 )){
328+ result. add(s. toString());
329+ }
330+ return ;
331+ }
332+ for (int i = startIndex; i < s. length(); i++ ){
333+ if (isValid(s, startIndex, i)){
334+ s. insert(i + 1 , ' .'
335+ backTracking(s, i + 2 , dotCount + 1 );
336+ s. deleteCharAt(i + 1 );
337+ }else {
338+ break ;
339+ }
340+ }
341+ }
342+ // [start, end]
343+ private boolean isValid (StringBuilder s , int start , int end ){
344+ if (start > end)
345+ return false ;
346+ if (s. charAt(start) == ' 0' && start != end)
347+ return false ;
348+ int num = 0 ;
349+ for (int i = start; i <= end; i++ ){
350+ int digit = s. charAt(i) - ' 0'
351+ num = num * 10 + digit;
352+ if (num > 255 )
353+ return false ;
354+ }
355+ return true ;
356+ }
357+ }
317358
318359// 方法二:比上面的方法时间复杂度低,更好地剪枝,优化时间复杂度
319360class Solution {
@@ -358,6 +399,7 @@ class Solution {
358399}
359400```
360401
402+ 361403## python
362404
363405python2:
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