216.组合总和III

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：依旧是正常遍历，过程中记录遍历的所有节点之和，如果当前元素之和已经大于所给定的值，退回上一节点

java：

class Solution {List<List<Integer>> result = new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();public List<List<Integer>> combinationSum3(int k, int n) {backTracking(n, k, 1, 0);return result;}private void backTracking(int targetSum, int k, int startIndex, int sum){if (sum > targetSum) {return;}if (path.size() == k) {if (sum == targetSum) result.add(new ArrayList<>(path));return;}for (int i = startIndex; i <= 9 - (k - path.size()) + 1; i++) {path.add(i);sum += i;backTracking(targetSum, k, i + 1, sum);path.removeLast();sum -= i;}}
}

17.电话号码的字母组合

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：依旧是组合，重点是拆分字符

java：

class Solution {List<String> list = new ArrayList<>();public List<String> letterCombinations(String digits) {if (digits == null || digits.length() == 0) {return list;}String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};backTracking(digits, numString, 0);return list;}StringBuilder temp = new StringBuilder();public void backTracking(String digits, String[] numString, int num) {if (num == digits.length()) {list.add(temp.toString());return;}String str = numString[digits.charAt(num) - '0'];for (int i = 0; i < str.length(); i++) {temp.append(str.charAt(i));backTracking(digits, numString, num + 1);temp.deleteCharAt(temp.length() - 1);}}
}