[leetcode]40. Combination Sum II

和39题不同的地方：
1.数组中有重复的数

2.数组中的数只能用一次

import java.util.*;

/**
 * Created by lvhao on 2017/7/3.
 * 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]
 ]
 相对于39题，由于每个数只能用一次，所以每次递归时遍历都是从index+1开始，而且由于数组中有重复的数，
 所以
 1.先用一个set去装结果，当set中不含有的时候才装进去，最后转化为list。
 2.加一个判断：当又回溯回来，本层的数要改变的时候，先判断是不是和前边得数相等，相等就continue
    但是怎么判断本次是不是回溯回来的呢：如果i >　index，就是。
 第二种方法更好
 */
public class Q40CombinationSum2 {
    public static void main(String[] args) {
        int[] nums = new int[]{10, 1, 2, 7, 6, 1, 5};
        System.out.println(combinationSum2(nums,8));
    }
    public static List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        List<List<Integer>> result = new ArrayList<>();
        backTracking(result,candidates,new ArrayList<>(),target,0);
        return new ArrayList<>(result);
    }
    public static void backTracking(List<List<Integer>> result,int[] candidates,
                                    List<Integer> cur,int left,int index)
    {
        if(left == 0)
        {
            List<Integer> temp = new ArrayList<>(cur);
            result.add(temp);
            return;
        }
        for(int i = index;i < candidates.length;i++)
        {
            if(i > index && candidates[i] == candidates[i-1])
                continue;
            if (candidates[i] <= left)
            {
                cur.add(candidates[i]);
                backTracking(result,candidates,cur,left-candidates[i],i+1);
                cur.remove(cur.size()-1);
            }
            else
                break;
        }

    }
}