求解k和问题-优快云博客

本文链接：https://blog.youkuaiyun.com/u013325815/article/details/103760348

本文介绍了一种使用回溯算法解决k和问题的方法，即在给定数组中找到所有可能的k个整数，使它们的和等于目标值。通过递归深度优先搜索，文章详细解释了如何遍历所有可能的组合并返回符合条件的解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given n unique postive integers, number k (1<=k<=n) and target.

Find all possible k integers where their sum is target.

Example

Example 1:

Input: [1,2,3,4], k = 2, target = 5
Output:  [[1,4],[2,3]]

Example 2:

Input: [1,3,4,6], k = 3, target = 8
Output:  [[1,3,4]]

思路：backtracking的标准模板；

public class Solution {
    /*
     * @param A: an integer array
     * @param k: a postive integer <= length(A)
     * @param target: an integer
     * @return: A list of lists of integer
     */
    public List<List<Integer>> kSumII(int[] A, int k, int target) {
        List<List<Integer>> lists = new ArrayList<List<Integer>>();
        if(A == null || A.length == 0 || k <= 0) {
            return lists;
        }
        
        List<Integer> list = new ArrayList<Integer>();
        dfs(A, k, target, list, lists, 0, 0, 0);
        return lists;
    }
    
    private void dfs(int[] A, 
                    int k, 
                    int target,
                    List<Integer> list,
                    List<List<Integer>> lists,
                    int index,
                    int curSum,
                    int count) {
        if(curSum > target) {
            return;
        }
        if(count == k && curSum == target) {
            lists.add(new ArrayList<Integer>(list));
            return;
        }    
        
        for(int i = index; i < A.length; i++) {
            list.add(A[i]);
            curSum += A[i];
            dfs(A, k, target, list, lists, i+1, curSum, count + 1);
            curSum -= A[i];
            list.remove(list.size()-1);
        }
    }
}

K Sum II

Example