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()) + 1
		for (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 2
        int 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);
        }
    }
}