【代码随想录Day27】

Day 27 回溯算法03

今日任务

• 39. 组合总和
• 40.组合总和II
• 131.分割回文串

代码实现

组合总和，直接套模板可解

 public List<List<Integer>> combinationSum(int[] candidates, int target) {backtracking(candidates, target, 0);return result;}void backtracking(int[] candidates, int target, int startIndex) {if (sum == target) {result.add(new ArrayList<>(path));return;}if (sum > target) {return;}for (int i = startIndex; i < candidates.length; i++) {path.add(candidates[i]);sum+=candidates[i];backtracking(candidates, target, i);sum-=candidates[i];path.remove(path.size() - 1);}}

组合总和II
这个就有点难，在模板之外要考虑怎么去重

public List<List<Integer>> combinationSum2(int[] candidates, int target) {Arrays.sort(candidates);backtracking2(candidates, target, 0, new boolean[candidates.length]);return result;}void backtracking2(int[] candidates, int target, int startIndex, boolean[] used) {if (sum == target) {result.add(new ArrayList<>(path));return;}if (sum > target) {return;}for (int i = startIndex; i < candidates.length; i++) {if (target - sum < candidates[i]) break;if ( i > 0 && candidates[i] == candidates[i - 1] && !used[i - 1]) {continue;}used[i] = true;path.add(candidates[i]);sum+=candidates[i];backtracking2(candidates, target, i + 1, used);used[i] = false;sum-=candidates[i];path.remove(path.size() - 1);}}

131.分割回文串
这个就太难了，模板还是那个模板，这里难以想到的是，当切割字符串的时候，从第二位开始，就相当于把第1、2位切割为一个字符，也就是说把startIndex当成切割线；至于解法中的动态规划部分，则完全不懂。

    public List<List<String>> partition(String s) {List<List<String>> result = new ArrayList<>();List<String> path = new ArrayList<>();backtracking(s, 0, result, path);return result;}void backtracking(String s, Integer startIndex, List<List<String>> result, List<String> path) {if (startIndex >= s.length()) {result.add(new ArrayList<>(path));return;}for (int i = startIndex; i < s.length(); i++) {String substring = s.substring(startIndex, i + 1);if (!isPalindrome(substring)) {continue;}path.add(substring);backtracking(s, i + 1, result, path);path.remove(path.size() - 1);}}public boolean isPalindrome(String s) {for (int i = 0; i < s.length()/2; i++) {if (s.charAt(i) != s.charAt(s.length() - i - 1)) {return false;}}return true;}

今日总结

1. 有点难，但是模板依然可用，每个题里边都有难以想象到的难点
2. 大数据？AI？有机会吗？
3. 今天+2，希望再来两天，2024就回本啦