17.电话号码的字母组合

在套模板的基础上，手动按位置放一个映射表，每次独立处理字符，还要对空字符串单独处理（因为默认生成了StringBuilder是空字符串而不是null）

此处for循环并不像之前从start开始遍历，因为本题每一个数字代表的是不同集合，也就是求不同集合之间的组合，而求组合问题是求同一个集合中的组合

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

39. 组合总和

这种复杂算法可以先把给的数组排序方便剪枝

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

40.组合总和II

candidates有重复元素，但还不能有重复的组合真的离谱，加一个数组来标记

class Solution {List<List<Integer>> res = new ArrayList<>();List<Integer> path = new ArrayList<>();boolean[] used;int sum = 0;public List<List<Integer>> combinationSum(int[] candidates, int target) {used = new boolean[candidates.length];Arrays.fill(used, false);Arrays.sort(candidates);backtrack(candidates, target, 0);return res;}void backtrack(int[] candidates, int target, int start) {if (sum == target) res.add(new ArrayList<>(path));     for (int i = start; i < candidates.length; i++) {if (sum + candidates[i] > target) break;if (i > 0 && candidates[i] == candidates[i - 1] && !used[i - 1]) continue;used[i] = true;path.add(candidates[i]);sum += candidates[i];backtrack(candidates, target, i+1);used[i] = false;path.removeLast();sum -= candidates[i];}}
}