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[原]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 ( a 1 , a 2 , … , a k ) must be in non-descending order. (ie, a 1 ≤ a 2 ≤ … ≤ a k ).
The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target ,

A solution set is:

[1, 7]

[1, 2, 5]

[2, 6]

[1, 1, 6]

思路分析：这题和Combination Sum很像，唯一的区别是每个数只能使用一次，只需要dfs传入start参数为当前位置的下一个位置i+1即可，详细分析参见Combination Sum的题解。两题代码几乎一样，只有一行不同。

AC Code

public class Solution {  public List<List<Integer>> combinationSum2(int[] num, int target) {   //1240   List<List<Integer>> res = new ArrayList<List<Integer>>();   int m = num.length;   if(m == 0) return  res;   Arrays.sort(num);   List<Integer> item = new ArrayList<Integer>();   dfs(num, target, 0, item, res);   return res;  }  public void dfs(int[] num, int target, int start, List<Integer> item, List<List<Integer>> res){   if(target == 0){    res.add(new ArrayList<Integer>(item));    return;   }   if(target < 0) return;   for(int i = start; i < num.length; i++){    if(i > start && num[i] == num[i-1]) continue;// avoid duplicate solutions    item.add(num[i]);    dfs(num, target - num[i], i + 1, item, res);    item.remove(item.size() - 1);   }  }  //1255 }

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