LeetCode Combination Sum II

标签：

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
}

LeetCode Combination Sum II

标签：

原文地址：http://blog.csdn.net/yangliuy/article/details/43835103