Java [Leetcode 40]Combination Sum II

题目描述：

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]

解题思路：

跟前一个类似，不过每次选好当前元素后，就直接选下一个位置的元素了。并且在每次选元素之前，保证这次的元素与上次的元素不重合。

代码如下：

public class Solution {
public List<List<Integer>> combinationSum2(int[] candidates,
int target) {
Arrays.sort(candidates);
List<List<Integer>> result = new ArrayList<List<Integer>>();
getResult(result, new ArrayList<Integer>(), candidates, target, 0);
return result;
}

public void getResult(List<List<Integer>> result,
List<Integer> current, int[] candiates, int target, int start) {
if (target > 0) {
for (int i = start; i < candiates.length && target >= candiates[i]; i++) {
if (i > start && candiates[i] == candiates[i - 1])
continue;
current.add(candiates[i]);
getResult(result, current, candiates, target - candiates[i],
i + 1);
current.remove(current.size() - 1);
}
} else if (target == 0) {
result.add(new ArrayList<Integer>(current));
}
}
}