216.组合总和III
题目链接
https://leetcode.cn/problems/combination-sum-iii/description/
题目描述
思路
自己写的效率会慢一些，而且没有用到剪枝
class Solution {List<List<Integer>> list = new ArrayList<>();List<Integer> list1 = new ArrayList<>();public List<List<Integer>> combinationSum3(int k, int n) {trackback(k,n,1);return list;}public void trackback(int k,int n,int startindex){if(list1.size()==k){int sum=0;for (Integer integer : list1) {sum+=integer;}if(sum==n){list.add(new ArrayList<>(list1));return;}}for (int i = startindex; i <= 9; i++) {list1.add(i);trackback(k,n,i+1);list1.remove(list1.size()-1);}}
}
参考答案
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) {// 减枝if (sum > targetSum) {return;}if (path.size() == k) {if (sum == targetSum) result.add(new ArrayList<>(path));return;}// 减枝 9 - (k - path.size()) + 1for (int i = startIndex; i <= 9 - (k - path.size()) + 1; i++) {path.add(i);sum += i;backTracking(targetSum, k, i + 1, sum);//回溯path.removeLast();//回溯sum -= i;}}
}
17.电话号码的字母组合
题目链接
https://leetcode.cn/problems/letter-combinations-of-a-phone-number/description/
题目描述
思路
class Solution {StringBuilder stringBuilder = new StringBuilder();List<String> list = new ArrayList<>();public List<String> letterCombinations(String digits) {if(digits == null || digits.length()==0){return list;}String[] nums = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};backTraing(digits,0,nums);return list;}public void backTraing(String digits,int index,String[] nums){//如果字符串是 “23”//当index等于2的时候，也就是已经遍历完了2 和 3，终止if(index==digits.length()){list.add(stringBuilder.toString());return;}//获取 2int digit = digits.charAt(index) - '0';//找到 2 对应的 "abc"String letter = nums[digit];for (int i = 0; i < letter.length(); i++) {//遍历 abc 添加到结果集中stringBuilder.append(letter.charAt(i));//回溯，将index 加 1 就会遍历 3 的 "def"backTraing(digits,index+1,nums);将添加的删掉stringBuilder.deleteCharAt(stringBuilder.length()-1);}}
}
