Combination Sum

Tags: Array, Backtracking, Medium

Question

Problem Statement

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

  • All numbers (including target) will be positive integers.
  • The solution set must not contain duplicate combinations.

Example 1:

  1. **Input:** candidates = [2,3,6,7], target = 7,
  2. **A solution set is:**
  3. [
  4.   [7],
  5.   [2,2,3]
  6. ]

Example 2:

  1. **Input:** candidates = [2,3,5], target = 8,
  2. **A solution set is:**
  3. [
  4.   [2,2,2,2],
  5.   [2,3,3],
  6.   [3,5]
  7. ]

题解

和 Permutations 十分类似，区别在于剪枝函数不同。这里允许一个元素被多次使用，故递归时传入的索引值不自增，而是由 for 循环改变。

Java

  1. public class Solution {
  2.     /**
  3.      * @param candidates: A list of integers
  4.      * @param target:An integer
  5.      * @return: A list of lists of integers
  6.      */
  7.     public List<List<Integer>> combinationSum(int[] candidates, int target) {
  8.         List<List<Integer>> result = new ArrayList<List<Integer>>();
  9.         List<Integer> list = new ArrayList<Integer>();
  10.         if (candidates == null) return result;
  12.         Arrays.sort(candidates);
  13.         helper(candidates, 0, target, list, result);
  15.         return result;
  16.     }
  18.     private void helper(int[] candidates, int pos, int gap,
  19.                         List<Integer> list, List<List<Integer>> result) {
  21.         if (gap == 0) {
  22.             // add new object for result
  23.             result.add(new ArrayList<Integer>(list));
  24.             return;
  25.         }
  27.         for (int i = pos; i < candidates.length; i++) {
  28.             // cut invalid candidate
  29.             if (gap < candidates[i]) {
  30.                 return;
  31.             }
  32.             list.add(candidates[i]);
  33.             helper(candidates, i, gap - candidates[i], list, result);
  34.             list.remove(list.size() - 1);
  35.         }
  36.     }
  37. }

源码分析

对数组首先进行排序是必须的，递归函数中本应该传入 target 作为入口参数，这里借用了 Soulmachine 的实现，使用 gap 更容易理解。注意在将临时 list 添加至 result 中时需要 new 一个新的对象。

复杂度分析

按状态数进行分析，时间复杂度 O(n!), 使用了list 保存中间结果，空间复杂度 O(n).

Reference

  • Soulmachine 的 leetcode 题解