leetcode40. Combination Sum II

40. Combination Sum II

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.
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]]

解法

在leetcode39. Combination Sum I的基础上改动了两处地方：

• backtrack(ret, temp, candidates, remain candidates[i], i + 1); //不包含本次元素

• if (i > start && candidates[i] == candidates[i- 1]) continue; // 去除重复的list，如[[1, 7], [1, 7]]，i>start为了避免数组越界，如0-1=-1

public class Solution {    public List<List<Integer>> combinationSum2(int[] candidates, int target) {        List<List<Integer>> ret = new ArrayList<>();        if (candidates == null || candidates.length == 0 || target <= 0) {            return ret;        }        Arrays.sort(candidates);        backtrack(ret, new ArrayList<Integer>(), candidates, target, 0);        return ret;    }    private void backtrack(List<List<Integer>> ret, List<Integer> temp, int[] candidates, int remain, int start) {        if (remain == 0) {            ret.add(new ArrayList<Integer>(temp));        } else if (remain < 0) {            return;        }        for (int i = start; i < candidates.length; i++) {            if (i > start && candidates[i] == candidates[i- 1]) continue;            temp.add(candidates[i]);            backtrack(ret, temp, candidates, remain - candidates[i], i + 1);            temp.remove(temp.size() - 1);        }    }}