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:

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

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代码

    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);        }    }