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代码随想录-035期-算法训练营【博客笔记汇总表】-CSDN博客
28 第七章 回溯算法
● 93.复原IP地址
● 78.子集
● 90.子集II 详细布置 93.复原IP地址 本期本来是很有难度的,不过 大家做完 分割回文串 之后,本题就容易很多了 题目链接/文章讲解:https://programmercarl.com/0093.%E5%A4%8D%E5%8E%9FIP%E5%9C%B0%E5%9D%80.html
视频讲解:https://www.bilibili.com/video/BV1XP4y1U73i/78.子集 子集问题,就是收集树形结构中,每一个节点的结果。 整体代码其实和 回溯模板都是差不多的。 题目链接/文章讲解:https://programmercarl.com/0078.%E5%AD%90%E9%9B%86.html
视频讲解:https://www.bilibili.com/video/BV1U84y1q7Ci 90.子集II 大家之前做了 40.组合总和II 和 78.子集 ,本题就是这两道题目的结合,建议自己独立做一做,本题涉及的知识,之前都讲过,没有新内容。 题目链接/文章讲解:https://programmercarl.com/0090.%E5%AD%90%E9%9B%86II.html
视频讲解:https://www.bilibili.com/video/BV1vm4y1F71J
往日任务
● day 1 任务以及具体安排:https://docs.qq.com/doc/DUG9UR2ZUc3BjRUdY
● day 2 任务以及具体安排:https://docs.qq.com/doc/DUGRwWXNOVEpyaVpG
● day 3 任务以及具体安排:https://docs.qq.com/doc/DUGdqYWNYeGhlaVR6
● day 4 任务以及具体安排:https://docs.qq.com/doc/DUFNjYUxYRHRVWklp
● day 5 周日休息
● day 6 任务以及具体安排:https://docs.qq.com/doc/DUEtFSGdreWRuR2p4
● day 7 任务以及具体安排:https://docs.qq.com/doc/DUElCb1NyTVpXa0Jj
● day 8 任务以及具体安排:https://docs.qq.com/doc/DUGdsY2JFaFhDRVZH
● day 9 任务以及具体安排:https://docs.qq.com/doc/DUHVXSnZNaXpVUHN4
● day 10 任务以及具体安排:https://docs.qq.com/doc/DUElqeHh3cndDbW1Q
●day 11 任务以及具体安排:https://docs.qq.com/doc/DUHh6UE5hUUZOZUd0
●day 12 周日休息
●day 13 任务以及具体安排:https://docs.qq.com/doc/DUHNpa3F4b2dMUWJ3
●day 14 任务以及具体安排:https://docs.qq.com/doc/DUHRtdXZZSWFkeGdE
●day 15 任务以及具体安排:https://docs.qq.com/doc/DUHN0ZVJuRmVYeWNv
●day 16 任务以及具体安排:https://docs.qq.com/doc/DUHBQRm1aSWR4T2NK
●day 17 任务以及具体安排:https://docs.qq.com/doc/DUFpXY3hBZkpabWFY
●day 18 任务以及具体安排:https://docs.qq.com/doc/DUFFiVHl3YVlReVlr
●day 19 周日休息
●day 20 任务以及具体安排:https://docs.qq.com/doc/DUGFRU2V6Z1F4alBH
●day 21 任务以及具体安排:https://docs.qq.com/doc/DUHl2SGNvZmxqZm1X
●day 22 任务以及具体安排:https://docs.qq.com/doc/DUHplVUp5YnN1bnBL
●day 23 任务以及具体安排:https://docs.qq.com/doc/DUFBUQmxpQU1pa29C
●day 24 任务以及具体安排:https://docs.qq.com/doc/DUEhsb0pUUm1WT2NP
●day 25 任务以及具体安排:https://docs.qq.com/doc/DUExTYXVzU1BiU2Zl
●day 26 休息
●day 27 任务以及具体安排:https://docs.qq.com/doc/DUElpbnNUR3hIbXlY
目录
0093_复原IP地址
0078_子集
0090_子集II
0093_复原IP地址
package com.question.solve.leetcode.programmerCarl2._08_backtrackingAlgorithms;import java.util.ArrayList;
import java.util.List;public class _0093_复原IP地址 {
}class Solution0093 {List<String> result = new ArrayList<>();public List<String> restoreIpAddresses(String s) {if (s.length() > 12) return result; // 算是剪枝了backTrack(s, 0, 0);return result;}//startIndex: 搜索的起始位置, pointNum:添加逗点的数量private void backTrack(String s, int startIndex, int pointNum) {if (pointNum == 3) {// 逗点数量为3时,分隔结束// 判断第四段⼦字符串是否合法,如果合法就放进result中if (isValid(s, startIndex, s.length() - 1)) {result.add(s);}return;}for (int i = startIndex; i < s.length(); i++) {if (isValid(s, startIndex, i)) {s = s.substring(0, i + 1) + "." + s.substring(i + 1); //在str的后⾯插⼊⼀个逗点pointNum++;backTrack(s, i + 2, pointNum);// 插⼊逗点之后下⼀个⼦串的起始位置为i+2pointNum--;// 回溯s = s.substring(0, i + 1) + s.substring(i + 2);// 回溯删掉逗点} else {break;}}}//判断字符串s在左闭⼜闭区间[start, end]所组成的数字是否合法private Boolean isValid(String s, int start, int end) {if (start > end) {return false;}if (s.charAt(start) == '0' && start != end) { // 0开头的数字不合法return false;}int num = 0;for (int i = start; i <= end; i++) {if (s.charAt(i) > '9' || s.charAt(i) < '0') { // 遇到⾮数字字符不合法return false;}num = num * 10 + (s.charAt(i) - '0');if (num > 255) { // 如果⼤于255了不合法return false;}}return true;}
}//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。
class Solution0093_2 {List<String> result = new ArrayList<>();public List<String> restoreIpAddresses(String s) {StringBuilder sb = new StringBuilder(s);backTracking(sb, 0, 0);return result;}private void backTracking(StringBuilder s, int startIndex, int dotCount) {if (dotCount == 3) {if (isValid(s, startIndex, s.length() - 1)) {result.add(s.toString());}return;}for (int i = startIndex; i < s.length(); i++) {if (isValid(s, startIndex, i)) {s.insert(i + 1, '.');backTracking(s, i + 2, dotCount + 1);s.deleteCharAt(i + 1);} else {break;}}}//[start, end]private boolean isValid(StringBuilder s, int start, int end) {if (start > end)return false;if (s.charAt(start) == '0' && start != end)return false;int num = 0;for (int i = start; i <= end; i++) {int digit = s.charAt(i) - '0';num = num * 10 + digit;if (num > 255)return false;}return true;}
}//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
class Solution0093_3 {List<String> result = new ArrayList<String>();StringBuilder stringBuilder = new StringBuilder();public List<String> restoreIpAddresses(String s) {restoreIpAddressesHandler(s, 0, 0);return result;}//number表示stringbuilder中ip段的数量public void restoreIpAddressesHandler(String s, int start, int number) {//如果start等于s的长度并且ip段的数量是4,则加入结果集,并返回if (start == s.length() && number == 4) {result.add(stringBuilder.toString());return;}//如果start等于s的长度但是ip段的数量不为4,或者ip段的数量为4但是start小于s的长度,则直接返回if (start == s.length() || number == 4) {return;}//剪枝:ip段的长度最大是3,并且ip段处于[0,255]for (int i = start; i < s.length() && i - start < 3 && Integer.parseInt(s.substring(start, i + 1)) >= 0&& Integer.parseInt(s.substring(start, i + 1)) <= 255; i++) {//如果ip段的长度大于1,并且第一位为0的话,continueif (i + 1 - start > 1 && s.charAt(start) - '0' == 0) {continue;}stringBuilder.append(s.substring(start, i + 1));//当stringBuilder里的网段数量小于3时,才会加点;如果等于3,说明已经有3段了,最后一段不需要再加点if (number < 3) {stringBuilder.append(".");}number++;restoreIpAddressesHandler(s, i + 1, number);number--;// 删除当前stringBuilder最后一个网段,注意考虑点的数量的问题stringBuilder.delete(start + number, i + number + 2);}}
}
0078_子集
package com.question.solve.leetcode.programmerCarl2._08_backtrackingAlgorithms;import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;public class _0078_子集 {
}class Solution0078 {List<List<Integer>> result = new ArrayList<>();// 存放符合条件结果的集合LinkedList<Integer> path = new LinkedList<>();// 用来存放符合条件结果public List<List<Integer>> subsets(int[] nums) {subsetsHelper(nums, 0);return result;}private void subsetsHelper(int[] nums, int startIndex) {result.add(new ArrayList<>(path));//「遍历这个树的时候,把所有节点都记录下来,就是要求的子集集合」。if (startIndex >= nums.length) { //终止条件可不加return;}for (int i = startIndex; i < nums.length; i++) {path.add(nums[i]);subsetsHelper(nums, i + 1);path.removeLast();}}
}class Solution0078_2 {List<Integer> t = new ArrayList<Integer>();List<List<Integer>> ans = new ArrayList<List<Integer>>();public List<List<Integer>> subsets(int[] nums) {int n = nums.length;for (int mask = 0; mask < (1 << n); ++mask) {t.clear();for (int i = 0; i < n; ++i) {if ((mask & (1 << i)) != 0) {t.add(nums[i]);}}ans.add(new ArrayList<Integer>(t));}return ans;}
}class Solution0078_3 {List<Integer> t = new ArrayList<Integer>();List<List<Integer>> ans = new ArrayList<List<Integer>>();public List<List<Integer>> subsets(int[] nums) {dfs(0, nums);return ans;}public void dfs(int cur, int[] nums) {if (cur == nums.length) {ans.add(new ArrayList<Integer>(t));return;}t.add(nums[cur]);dfs(cur + 1, nums);t.remove(t.size() - 1);dfs(cur + 1, nums);}
}
0090_子集II
package com.question.solve.leetcode.programmerCarl2._08_backtrackingAlgorithms;import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;public class _0090_子集II {
}class Solution0090 {List<List<Integer>> res = new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();public List<List<Integer>> subsetsWithDup(int[] nums) {Arrays.sort(nums);backTrack(nums, 0);return res;}private void backTrack(int[] nums, int startIndex) {if (!res.contains(new ArrayList<>(path))) {res.add(new ArrayList<>(path));}for (int i = startIndex; i < nums.length; i++) {path.add(nums[i]);backTrack(nums, i + 1);path.removeLast();}}
}class Solution0090_2 {//使用used数组List<List<Integer>> result = new ArrayList<>();//存放符合条件结果的集合LinkedList<Integer> path = new LinkedList<>(); //用来存放符合条件结果boolean[] used;public List<List<Integer>> subsetsWithDup(int[] nums) {if (nums.length == 0) {result.add(path);return result;}Arrays.sort(nums);used = new boolean[nums.length];subsetsWithDupHelper(nums, 0);return result;}private void subsetsWithDupHelper(int[] nums, int startIndex) {result.add(new ArrayList<>(path));if (startIndex >= nums.length) {return;}for (int i = startIndex; i < nums.length; i++) {if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) {continue;}path.add(nums[i]);used[i] = true;subsetsWithDupHelper(nums, i + 1);path.removeLast();used[i] = false;}}
}class Solution0090_3 {//不使用used数组List<List<Integer>> res = new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();public List<List<Integer>> subsetsWithDup(int[] nums) {Arrays.sort(nums);subsetsWithDupHelper(nums, 0);return res;}private void subsetsWithDupHelper(int[] nums, int start) {res.add(new ArrayList<>(path));for (int i = start; i < nums.length; i++) {// 跳过当前树层使用过的、相同的元素if (i > start && nums[i - 1] == nums[i]) {continue;}path.add(nums[i]);subsetsWithDupHelper(nums, i + 1);path.removeLast();}}
}
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