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 (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] 

NP问题，遇到很多次了，这道题跟Combination Sum很像

第二遍方法：注意17行的跳过条件, 我在这里做错过，i > starter 才行， 仅仅i > 0是不行的。 i > starter限制的是一个数的取值不要重复，比如5,5,7,8，第一个数已经取过第一个5了，它就不要再尝试第二个5，但第二个数可以去尝试第二个5，所以第一个数取5同时第二个数也取5是可以的。但是如果把条件写成i > 0，第一个数取了第一个5，第二个数就连第二个5都取不了了，就没有第一个取5第二个也取5这种情况了

 1 public class Solution {

 2     public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {

 3         ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();

 4         ArrayList<Integer> item = new ArrayList<Integer>();

 5         Arrays.sort(num);

 6         helper(res, item, num, target, 0);

 7         return res;

 8     }

 9     

10     public void helper(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> item, int[] num, int remain, int starter) {

11         if (remain < 0) return;

12         if (remain == 0) {

13             res.add(new ArrayList<Integer>(item));

14             return;

15         }

16         for (int i=starter; i<num.length; i++) {

17             if (i>starter && num[i] == num[i-1]) continue;

18             item.add(num[i]);

19             helper(res, item, num, remain-num[i], i+1);

20             item.remove(item.size()-1);

21         }

22     }

23 }

Naive方法：

 1 public class Solution {

 2     public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {

 3         ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();

 4         if (num == null || num.length == 0) return res;

 5         ArrayList<Integer> path = new ArrayList<Integer>();

 6         Arrays.sort(num);

 7         helper(res, path, num, target, 0, 0);

 8         return res;

 9     }

10     

11     public void helper(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> path, int[] num, int target, int accum, int index) {

12         if (accum > target) return;

13         if (accum == target) {

14             if (!res.contains(path)) res.add(new ArrayList<Integer>(path));

15             return;

16         }

17         for (int i=index; i<num.length; i++) {

18             path.add(num[i]);

19             helper(res, path, num, target, accum+num[i], i+1);

20             path.remove(path.size()-1);

21         }

22     }

23 }

CodeGanker的做法：注意21行他处理重复的情况，直接跳过

 1 public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {

 2     ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();

 3     if(num == null || num.length==0)

 4         return res;

 5     Arrays.sort(num);

 6     helper(num,0,target,new ArrayList<Integer>(),res);

 7     return res;

 8 }

 9 private void helper(int[] num, int start, int target, ArrayList<Integer> item,

10 ArrayList<ArrayList<Integer>> res)

11 {

12     if(target == 0)

13     {

14         res.add(new ArrayList<Integer>(item));

15         return;

16     }

17     if(target<0 || start>=num.length)

18         return;

19     for(int i=start;i<num.length;i++)

20     {

21         if(i>start && num[i]==num[i-1]) continue;

22         item.add(num[i]);

23         helper(num,i+1,target-num[i],item,res);

24         item.remove(item.size()-1);

25     }

26 }