【代码随想录Day28】

Day 28 回溯 Part04

今日任务

93.复原IP地址
78.子集
90.子集II

代码实现

今天的题出乎意料的简单
复原IP地址，和昨天的分割回文串比较像
一个是分割，一个是判断ip有效

    List<String> result = new ArrayList<>();List<String> path = new ArrayList<>();public List<String> restoreIpAddresses(String s) {if (s.length() < 4 || s.length() > 12) {return result;}backtracking(s, 0);return result;}void backtracking(String s, Integer startIndex) {if (path.size() == 4) {if (startIndex == s.length()) {result.add(String.join(".", path));}return;}for (int i = startIndex; i < s.length() && i < startIndex + 3; i++) {String substring = s.substring(startIndex, i + 1);if (substring.length() > 1 && substring.charAt(0) == '0') {return;}if (Integer.parseInt(substring) > 255) {return;}int remainLength = s.substring(i).length();if (remainLength < 4 - path.size() || remainLength > 3 * (4 - path.size())) {return;}path.add(substring);backtracking(s, i + 1);path.remove(path.size() - 1);}}

子集问题，完完全全的模板

List<List<Integer>> res = new ArrayList<>();List<Integer> path1 = new ArrayList<>();public List<List<Integer>> subsets(int[] nums) {backtracking(nums, 0);return res;}void backtracking(int[] nums, int startIndex) {res.add(new ArrayList<>(path1));if (startIndex >= nums.length) {return;}for (int i = startIndex; i < nums.length; i++) {path1.add(nums[i]);backtracking(nums, i + 1);path1.remove(path1.size() - 1);}}

子集II，比子集多了一个去重，然后去重是昨天做过的

public List<List<Integer>> subsetsWithDup(int[] nums) {Arrays.sort(nums);boolean[] used = new boolean[nums.length];backtracking(nums, 0, used);return res;}void backtracking(int[] nums, int startIndex, boolean[] used) {res.add(new ArrayList<>(path1));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;}path1.add(nums[i]);used[i] = true;backtracking(nums, i + 1, used);path1.remove(path1.size() - 1);used[i] = false;}}