Day28 回溯算法 part04

93.复原IP地址

我的思路：
在分割回文串代码的基础上改写，需要考虑多种情况
不合法的IP地址part：长度>1同时首数字为0 || 长度=0 || 长度＞3 || ＞255
将各个part结合起来添加 ”." 的时候，需要分前三part和最后一个，因此回溯完删除的范围也需要分别考虑

class Solution {public List<String> res = new ArrayList();public StringBuffer path = new StringBuffer();public List<String> restoreIpAddresses(String s) {if(s == null || s.length() == 0) {return new ArrayList();}backtracing(s, 0, 0);return res;}public void backtracing(String s, int startIndex, int partNum) {if(partNum == 4) {if(startIndex == s.length()) {StringBuffer sb = new StringBuffer(path);res.add(sb.toString());}return;}for(int i = startIndex; i < s.length(); i++) {String part = s.substring(startIndex, i + 1);if(isValid(part)) {if(partNum < 3) {path.append(part);path.append('.');backtracing(s, i + 1 , partNum + 1);path.delete(path.length() - part.length() - 1, path.length());}else {path.append(part);backtracing(s, i + 1, partNum + 1);path.delete(path.length() - part.length(), path.length());}  }else {break;}}}public boolean isValid(String s) {if(s.length() == 0 || s.length() > 3 || (s.charAt(0) == '0' && s.length() > 1) || Integer.parseInt(s) > 255) {return false;}return true;}
}

78.子集

我的思路：
不需要考虑返回情况，将path全部加到result集合中

解答：

class Solution {public List<Integer> path = new ArrayList();public List<List<Integer>> res = new ArrayList();public List<List<Integer>> subsets(int[] nums) {Arrays.sort(nums);backtracing(nums, 0);return res;}public void backtracing(int[] nums, int startIndex) {res.add(new ArrayList(path));for(int i = startIndex; i < nums.length; i++) {path.add(nums[i]);backtracing(nums, i+1);path.remove(path.size() - 1);}}
}

90.子集II

我的思路：
在上一题的基础上，判断要添加到path，在result里面是不是已经存在
另外模仿组合的做法进行了剪枝

解答:

class Solution {public List<Integer> path = new ArrayList();public List<List<Integer>> res = new ArrayList();public List<List<Integer>> subsetsWithDup(int[] nums) {Arrays.sort(nums);backtracing(nums, 0);return res;}public void backtracing(int[] nums, int startIndex) {if(hasNums(path, res)) {res.add(new ArrayList(path));}for(int i = startIndex; i < nums.length; i++) {if(i > startIndex && nums[i] == nums[i-1]) {continue;}path.add(nums[i]);backtracing(nums, i+1);path.remove(path.size() - 1);}}public boolean hasNums(List<Integer> path, List<List<Integer>> res) {for(List l : res) {if(path.equals(l)){return false;}}return true;}
}