DAY28| 93. 复原IP地址 ,79.子集 ,90.子集II

本文主要是介绍DAY28| 93. 复原IP地址 ,79.子集 ,90.子集II，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

文章目录

- 93.复原IP地址
- 78.子集
- 90.子集II

93.复原IP地址

文字讲解：复原IP地址

视频讲解：复原IP地址

**状态：**此题调试了几次ok，与昨天的分割回文子串相比，就是在判断终止条件处需要处理；

思路：

代码：

class Solution {List<String> result = new ArrayList<>();LinkedList<String> tempList = new LinkedList<>();public List<String> restoreIpAddresses(String s) {backTracking(0, s);return result;}public void backTracking(Integer startIndex, String s) {if (tempList.size() == 3) {if (validIpParam(s.substring(startIndex, s.length()))) {tempList.offer(s.substring(startIndex, s.length()));result.add(getIp());tempList.pollLast();}return;}for (int i = startIndex; i < s.length() && tempList.size()<=4; i++) {if (validIpParam(s.substring(startIndex, i+1))) {tempList.offer(s.substring(startIndex, i+1));} else {continue;}backTracking(i+1, s);tempList.pollLast();}}public boolean validIpParam(String s) {if (s == null || s.length()==0) {return false;}if (s.length()>=2 && s.charAt(0)=='0') {return false;}if (s.length()>3) {return false;}Integer num = Integer.valueOf(s);if (num < 0 || num > 255) {return false;}return true;}public String getIp() {StringBuilder resultStr = new StringBuilder();for (int i = 0; i < tempList.size(); i++) {if (i==tempList.size()-1) {resultStr.append(tempList.get(i));} else {resultStr.append(tempList.get(i)).append(".");}}return resultStr.toString();}
}

78.子集

文字讲解：子集

视频讲解：子集

状态：这一题的关键在于收集元素的位置，理解了回溯算法中的树形结构和理论知识，这题可以想到在for循环中收集元素即可

思路：

代码：

class Solution {List<List<Integer>> result = new ArrayList<>();LinkedList<Integer> tempList = new LinkedList<>();public List<List<Integer>> subsets(int[] nums) {backTracking(nums, 0);result.add(new ArrayList<>());return result;}public void backTracking(int[] nums, int index) {if (index>=nums.length) {return;}for (int i = index; i < nums.length; i++) {tempList.add(nums[i]);//收集元素result.add(new ArrayList<>(tempList));backTracking(nums, i+1);tempList.pollLast();}}
}

90.子集II

文字讲解：子集II

视频讲解：子集II

状态：这题秒了

思路：

代码：

class Solution {List<List<Integer>> result = new ArrayList<>();LinkedList<Integer> tempList = new LinkedList<>();public List<List<Integer>> subsetsWithDup(int[] nums) {result.add(tempList);//对数组先进行排序Arrays.sort(nums);backTracking(nums, 0);return result;}public void backTracking(int[] nums, int startIndex) {if (startIndex>=nums.length) {return;}for (int i = startIndex; i < nums.length; i++) {if (i>startIndex&&nums[i]==nums[i-1]) {continue;}tempList.offer(nums[i]);result.add(new ArrayList<>(tempList));backTracking(nums, i+1);tempList.pollLast();}}
}