LeetCode 40. 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.
• 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]]

本来想着用Set来去除重，结果TLE啦。那只能在递归中的for循环里面去重了。

public class Solution {    private void helper(int[] candidates, int pos, int target, Stack<Integer> stack, List<List<Integer>> list){        if(target < 0) return ;        if(target == 0){            List<Integer> row = new ArrayList<>();            for(int i:stack){                row.add(i);            }            list.add(row);            return;        }        int prev = 0;        for(int i=pos; i<candidates.length; i++){            if(candidates[i] == prev) continue;            prev =candidates[i];            stack.push(candidates[i]);            helper(candidates, i+1, target-candidates[i], stack, list);            stack.pop();        }            }    public List<List<Integer>> combinationSum2(int[] candidates, int target) {        Stack<Integer> stack = new Stack<>();        List<List<Integer>> lists = new ArrayList<>();        Arrays.sort(candidates);        helper(candidates, 0, target, stack, lists);        return lists;    }}