Combination Sum II 153

Question

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.

Notice

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.

Example

Given candidate set [10,1,6,7,2,1,5] and target 8,

A solution set is:

[ [1,7],

[1,2,5],

[2,6],

[1,1,6] ]

Solution

与Combination Sum类似。但是因为每个数只能用一次，下一层pos为上一层pos＋1。避免重复的方法：重复的数只能取第一个元素。

代码如下：

需要额外O(n)空间记录元素状态

public class Solution {
    /**
     * @param num: Given the candidate numbers
     * @param target: Given the target number
     * @return: All the combinations that sum to target
     */
    public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {
        // write your code here
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(num == null || num.length == 0 || target <= 0){
            return result;
        }

        ArrayList<Integer> list = new ArrayList<Integer>();
        Arrays.sort(num);
        int[] visit = new int[num.length];
        helper(result, list, num, target, 0, visit);

        return result;
    }

    private void helper(ArrayList<ArrayList<Integer>> result, ArrayList<Integer> list, int[] num, int target, int pos, int[] visit){
        if(target == 0){
            result.add(new ArrayList<Integer>(list));
            return;
        }

        for(int i = pos; i < num.length; i++){
            if(visit[i] == 1 || (i != 0 && num[i] == num[i - 1] && visit[i - 1]== 0)){
                continue;
            }

            if(num[i] > target){
                return;
            }
            list.add(num[i]);
            visit[i] = 1;
            helper(result, list, num, target - num[i], i + 1, visit);
            list.remove(list.size() - 1);
            visit[i] = 0;
        }
    }
}

Revised version:

public class Solution {

    private ArrayList<ArrayList<Integer>> results;

    public ArrayList<ArrayList<Integer>> combinationSum2(int[] candidates,
            int target) {
        if (candidates.length < 1) {
            return results;
        }

        ArrayList<Integer> path = new ArrayList<Integer>();
        java.util.Arrays.sort(candidates);
        results = new ArrayList<ArrayList<Integer>> ();
        combinationSumHelper(path, candidates, target, 0);

        return results;
    }

    private void combinationSumHelper(ArrayList<Integer> path, int[] candidates, int sum, int pos) {
        if (sum == 0) {
            results.add(new ArrayList<Integer>(path));
        }

        if (pos >= candidates.length || sum < 0) {
            return;
        }

        int prev = -1;
        for (int i = pos; i < candidates.length; i++) {
            //重复的数只能取第一个元素
            if (candidates[i] != prev) {
                path.add(candidates[i]);
                combinationSumHelper(path, candidates, sum - candidates[i], i + 1);
                prev = candidates[i];
                path.remove(path.size()-1);
            }
        }
    }

}