39. 组合总和

题目链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

解题思路：candidate的元素个数为n,从n的n次方中选出可能的组合，大于目标则抛弃该组合，进行回溯

java：

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

40.组合总和II

题目链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

解题思路：先排序，然后遍历，如果当前结点与上一节点相同，则将会出现重复的遍历结果

class Solution {public List<List<Integer>> combinationSum2(int[] candidates, int target) {List<List<Integer>> res = new ArrayList<>();Arrays.sort(candidates); backtracking(res, new ArrayList<>(), candidates, target, 0, 0);return res;}public void backtracking(List<List<Integer>> res, List<Integer> path, int[] candidates, int target, int sum, int idx) {if (sum == target) {res.add(new ArrayList<>(path));return;}for (int i = idx; i < candidates.length; i++) {if (sum + candidates[i] > target) break;if(i>idx&&candidates[i]==candidates[i-1]) continue;path.add(candidates[i]);backtracking(res, path, candidates, target, sum + candidates[i], i+1);path.remove(path.size() - 1); }}
}

131.分割回文串

题目链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

java：

class Solution {List<List<String>> lists = new ArrayList<>();Deque<String> deque = new LinkedList<>();public List<List<String>> partition(String s) {backTracking(s, 0);return lists;}private void backTracking(String s, int startIndex) {if (startIndex >= s.length()) {lists.add(new ArrayList(deque));return;}for (int i = startIndex; i < s.length(); i++) {if (isPalindrome(s, startIndex, i)) {String str = s.substring(startIndex, i + 1);deque.addLast(str);} else {continue;}backTracking(s, i + 1);deque.removeLast();}}private boolean isPalindrome(String s, int startIndex, int end) {for (int i = startIndex, j = end; i < j; i++, j--) {if (s.charAt(i) != s.charAt(j)) {return false;}}return true;}
}