leetCode 40.Combination Sum II(组合总和II) 解题思路和方法

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] 

思路：此题和Combination Sum类同，而且很多代码都可以复用，仅仅改变几处不大的变化和增加了一个去处重复项的函数。

具体代码如下：

public class Solution {

    public List> combinationSum2(int[] a, int t) {
		List> list = new ArrayList>();
		List> newList = new ArrayList>();
		list = combinationSum1(a,t);
        Set> set = new HashSet>();//去除重复
        for(int i = 0; i < list.size();i++){
        	if(set.add(list.get(i))){//没有重复
        		newList.add(list.get(i));//添加新的列表
        	}
        }
        return newList;
	}
	
    public List> combinationSum1(int[] a, int t) {
        List> list = new ArrayList>();
        Arrays.sort(a);//数组排序
        //各种特殊情况
        if(a.length == 0 || a[0] > t)
            return list;

        int len = 0;
        boolean isAdd = false;//控制与t相等的数只添加一次
        for(int i = 0; i< a.length;i++){
            if(a[i] == t){
                if(!isAdd){//如果未添加
                    List al = new ArrayList();
                    al.add(t);
                    list.add(al);
                    isAdd = true;//标记已添加
                }
            }else if(a[i] < t){//只要比target小的值，大的值肯定不满足，排除
                a[len++] = a[i];//新数组
            }
        }
        //只存在a[0] < target 或 a[0] > target
        if(a[0] > t)//肯定已没有符合要求的组合
            return list;
        //a[0] < target
        
        for(int i = 0; i < len; i++){//循环搜索符合要求的数字组合
        	int[] b = Arrays.copyOfRange(a, i+1, len);//不含>=t数据的新数组
        	if(a[i] > t)//如果a[i]，肯定以后的数据已不符合，返回
        		break;
            //相等于已经有了一个值a[0]了    
            List> newList = new ArrayList>();
            if(i < len -1)
            	newList = combinationSum1(b,t-a[i]);
            if(newList.size() > 0){//里面有符合要求的数据
                for(int j = 0; j < newList.size();j++){
                    newList.get(j).add(a[i]);//新返回的各个值添加a[0]
                	Collections.sort(newList.get(j));//排序
                    list.add(newList.get(j));//最终添加
                }
            }
        }
        return list;
    }
}