40. Combination Sum II

Problem

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates may only be used once in the combination.

Note:

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

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

Solution (Java)

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> sums = new ArrayList<>();
        // optimize
        Arrays.sort(candidates);
        combinationSum(candidates, target, 0, sums, new LinkedList<>());
        return sums;
    }

    private void combinationSum(
            int[] candidates,
            int target,
            int start,
            List<List<Integer>> sums,
            LinkedList<Integer> sum) {
        if (target == 0) {
            // make a deep copy of the current combination
            sums.add(new ArrayList<>(sum));
            return;
        }
        for (int i = start; i < candidates.length && target >= candidates[i]; i++) {
            // If candidate[i] equals candidate[i-1], then solutions for i is subset of
            // solution of i-1
            if (i == start || (i > start && candidates[i] != candidates[i - 1])) {
                sum.addLast(candidates[i]);
                // call on 'i+1' (not i) to avoid duplicate usage of same element
                combinationSum(candidates, target - candidates[i], i + 1, sums, sum);
                sum.removeLast();
            }
        }
    }
}

Solution (Javascript)

/**
 * @param {number[]} candidates
 * @param {number} target
 * @return {number[][]}
 */
var combinationSum2 = function(candidates, target) {
  var res = [];
  var len = candidates.length;
  candidates.sort((a, b) => (a - b));
  dfs(res, [], 0, len, candidates, target);
  return res;
};

var dfs = function (res, stack, index, len, candidates, target) {
  var tmp = null;
  if (target < 0) return;
  if (target === 0) return res.push(stack);
  for (var i = index; i < len; i++) {
    if (candidates[i] > target) break;
    if (i > index && candidates[i] === candidates[i - 1]) continue;
    tmp = Array.from(stack);
    tmp.push(candidates[i]);
    dfs(res, tmp, i + 1, len, candidates, target - candidates[i]);
  }
};

Explain:

与之前一题不同的地方是：

1. 候选数字可能有重复的
2. 单个候选数字不能重复使用

Complexity:

• Time complexity : O(n^2).
• Space complexity : O(n^2).