leeetcode-40. Combination Sum II

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]]

题解：

和之前的那个Combination Sum都是一脉相承的

一开始我看到问题，只是看到了元素不能重复取，所以就把前一题的递归中的 i 改成了 i + 1

但是提交完，发现还是wrong answer，原来给定的集合里的元素是可以重复的，就导致结果中的集合就会重复，所以还要加一个判定条件，当有相同的元素时，只使用最后一个元素并入递归就好：

错误的例子：

Input:[10,1,2,7,6,1,5]8

Output:[[1,1,6],[1,2,5],[1,7],[1,2,5],[1,7],[2,6]]

Expected:[[1,1,6],[1,2,5],[1,7],[2,6]]

AC的solu:

class Solution {      public List<List<Integer>> combinationSum2(int[] nums, int target) {          List<List<Integer>> list = new ArrayList<>();          Arrays.sort(nums);          backtrack(list, new ArrayList<>(), nums, target, 0);          return list;      }        private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){          if(remain < 0) return;          else if(remain == 0) list.add(new ArrayList<>(tempList));          else{             for(int i = start; i < nums.length; i++){                  if(i != start && nums[i] == nums[i-1])                    continue;                tempList.add(nums[i]);                  backtrack(list, tempList, nums, remain - nums[i], i+1); // not i + 1 because we can reuse same elements                  tempList.remove(tempList.size() - 1);              }          }      }  }