leetcode216. Combination Sum III

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最新推荐文章于 2025-08-22 19:21:44 发布

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分类专栏： leetcode 文章标签： leetcode backtrack array

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本文介绍了一种寻找所有可能的组合方式，使得k个数相加等于目标数n的问题解决方法。仅使用1到9的数字，每组组合内的数字各不相同。通过两种回溯法实现，提供清晰的代码示例。

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216. Combination Sum III

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]

Example 2:

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

解法

回溯法，有序的，不包括本元素，并且判断条件增加temp.size() == num

public class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        int[] candidates = new int[9];
        for (int i = 0; i < candidates.length; i++) {
            candidates[i] = i + 1;
        }
        List<List<Integer>> ret = new ArrayList<>();
        if (candidates == null || candidates.length == 0 || n <= 0) {
            return ret;
        }
        Arrays.sort(candidates);
        backtrack(ret, new ArrayList<Integer>(), candidates, n, k, 0);
        return ret;
    }
    private void backtrack(List<List<Integer>> ret, List<Integer> temp, int[] candidates, int remain, int num, int start) {
        if (remain == 0 && temp.size() == num) {
            ret.add(new ArrayList<Integer>(temp));
        } else if (remain < 0 || temp.size() > num) {
            return;
        }
        for (int i = start; i < candidates.length; i++) {
            if (candidates[i] == candidates[i] - 1) continue;
            temp.add(candidates[i]);
            backtrack(ret, temp, candidates, remain - candidates[i], num, i + 1);
            temp.remove(temp.size() - 1);
        }
    }
}

解法二

解法一的优化

public class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        List<List<Integer>> ret = new ArrayList<>();
        backtrack(ret, new ArrayList<Integer>(), n, k, 1);
        return ret;
    }
    private void backtrack(List<List<Integer>> ret, List<Integer> temp, int remain, int num, int start) {
        if (remain == 0 && temp.size() == num) {
            ret.add(new ArrayList<Integer>(temp));
        } else if (remain < 0 || temp.size() > num) {
            return;
        }
        for (int i = start; i <= 9; i++) {
            temp.add(i);
            backtrack(ret, temp, remain - i, num, i + 1);
            temp.remove(temp.size() - 1);
        }
    }
}