Combination Sum II

Question

  1. Given a collection of candidate numbers (C) and a target number (T),
  2. find all unique combinations in C where the candidate numbers sums to T.
  3. Each number in C may only be used once in the combination.
  5. Have you met this question in a real interview? Yes
  6. Example
  7. For example, given candidate set 10,1,6,7,2,1,5 and target 8,
  9. A solution set is:
  11. [1,7]
  13. [1,2,5]
  15. [2,6]
  17. [1,1,6]
  19. Note
  20. All numbers (including target) will be positive integers.
  21. Elements in a combination (a1, a2, … , ak) must be in non-descending order.
  22. (ie, a1 ≤ a2 ≤ … ≤ ak).
  23. The solution set must not contain duplicate combinations.

题解

和 Unique Subsets 非常类似。在 Combination Sum 的基础上改改就好了。

Java

  1. public class Solution {
  2.     /**
  3.      * @param num: Given the candidate numbers
  4.      * @param target: Given the target number
  5.      * @return: All the combinations that sum to target
  6.      */
  7.     public List<List<Integer>> combinationSum2(int[] num, int target) {
  8.         List<List<Integer>> result = new ArrayList<List<Integer>>();
  9.         List<Integer> list = new ArrayList<Integer>();
  10.         if (num == null) return result;
  12.         Arrays.sort(num);
  13.         helper(num, 0, target, list, result);
  15.         return result;
  16.     }
  18.     private void helper(int[] nums, int pos, int gap,
  19.                         List<Integer> list, List<List<Integer>> result) {
  21.         if (gap == 0) {
  22.             result.add(new ArrayList<Integer>(list));
  23.             return;
  24.         }
  26.         for (int i = pos; i < nums.length; i++) {
  27.             // ensure only the first same num is chosen, remove duplicate list
  28.             if (i != pos && nums[i] == nums[i - 1]) {
  29.                 continue;
  30.             }
  31.             // cut invalid num
  32.             if (gap < nums[i]) {
  33.                 return;
  34.             }
  35.             list.add(nums[i]);
  36.             // i + 1 ==> only be used once
  37.             helper(nums, i + 1, gap - nums[i], list, result);
  38.             list.remove(list.size() - 1);
  39.         }
  40.     }
  41. }

源码分析

这里去重的方法继承了 Unique Subsets 中的做法，当然也可以新建一变量 prev，由于这里每个数最多只能使用一次，故递归时索引变量传i + 1.

复杂度分析

时间复杂度 O(n), 空间复杂度 O(n).