LeetCode 040 Combination Sum II

题目描述

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

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

Note:

All numbers (including target) will be positive integers.

Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).

The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8,

A solution set is:

[code][1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6]

代码

[code]    Set<List<Integer>> result;
    ArrayList<Integer> cur;
    int[] candidates;
    int target;

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {

        if (candidates == null || candidates.length == 0) {
            return new ArrayList<List<Integer>>();
        }

        result = new HashSet<List<Integer>>();
        cur = new ArrayList<Integer>();

        this.candidates = candidates;
        this.target = target;

        Arrays.sort(candidates);
        dfs(0, target);

        return new ArrayList<List<Integer>>(result);
    }

    void dfs(int j, int target) {

        // 找到解，存到result中
        if (target == 0) {
            result.add(new ArrayList<Integer>(cur));
            return;
        }

        for (int i = j; i < candidates.length; i++) {

            // 如果小于就返回，表明此后不会有解了！
            if (candidates[i] > target) {
                return;
            }

            // 递归求解
            cur.add(candidates[i]);
            dfs(i + 1, target - candidates[i]);
            cur.remove(cur.size() - 1);
        }
    }