LeetCode 40. Combination Sum II （组合的和之二）

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]
]

 

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题目标签：Array

　　这道题目和前面那一题本质一摸一样，不同之处是，这题每一个数字只能用一次，之前那题可以无限用；这一题给的array里有重复的，之前那题没有。所以只要之前那题做了，稍微改动一下就可以了。把递归下去的数字index 从i 改成 i + 1，因为这里不需要重复的数字了。再有就是要设一个条件，如果一个重复的数字，不是出现在第一位的话，就跳过它。

 

 

Java Solution:

Runtime beats 83.86% 

完成日期：07/17/2017

关键词：Array

关键点：Backtracking with sorted array

 

public class Solution 
{
    public List<List<Integer>> combinationSum2(int[] candidates, int target) 
    {
        List<List<Integer>> list = new ArrayList<>();
        Arrays.sort(candidates);
        backtrack(list, new ArrayList<>(), candidates, target, 0);
        
        return list;
    }
    
    public boolean backtrack(List<List<Integer>> list, List<Integer> tempList, 
            int[] nums, int remain, int start)
    {
        if(remain < 0) // if remain is 0 or less than 0, meaning the rest numbers are even greater
            return false; // therefore, no need to continue the loop, return false
        else if(remain == 0)
        {
            list.add(new ArrayList<>(tempList));
            return false;
        }
        else
        {
            for(int i=start; i<nums.length; i++)
            {
                if(i > start && nums[i] == nums[i-1])
                    continue;
                boolean flag;
                tempList.add(nums[i]);
                flag = backtrack(list, tempList, nums, remain - nums[i], i+1); // i + 1 because we cannot use same number.
                tempList.remove(tempList.size() - 1);
                
                if(!flag) // if find a sum or fail to find a sum, there is no need to continue
                    break;// because it is a sorted array with no duplicates, the rest numbers are even greater.
            }
            
            return true; // return true because previous tempList didn't find a sum or fail a sum
        }

    }
}

参考资料：

http://www.cnblogs.com/grandyang/p/4419386.html

 

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