LeetCode 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.

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]

题意分析：从给定数组中找到一组数字，要求这组数字之和等于target。另外，数组中的数字不允许被使用多次，但如果一开始就存在多个的话，可以使用多次。

解题思路：显然先排序，然后dfs。其中有一点要注意的是：因为不能重复，所以要跳过一样的数字。以上面为例，如果不跳过重复的1的话，会出现多个：[1,7]。

上AC代码：

public class Solution {
    
    void dfs(int [] num, int start, int target, ArrayList<Integer> array, ArrayList<ArrayList<Integer>> result) {
        if(target==0) {
            result.add(new ArrayList<Integer>(array));
            return;
        }
        
        if(start>=num.length||num[0]>target) {
            return;
        }
        int i = start;
        while(i<num.length) {
            if(num[i]<=target) {
                array.add(num[i]);
                dfs(num, i + 1, target - num[i], array, result);
                array.remove(array.size()-1);
                //跳过重复的元素
                while(i<(num.length-1)&&num[i]==num[i+1]) {
                    i++;
                }
            }
            i++;
        }
    }
    
    
    public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {
         ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
         ArrayList<Integer> array = new ArrayList<Integer>();
         if(num==null) {
             result.add(array);
             return result;
         }
         Arrays.sort(num);
         dfs(num,0, target,array,result); 
         return result;
    }
}