小白笔记----------------------------------------leetcode(40. Combination Sum II )

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations inC 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 target8,
A solution set is:

[  [1, 7],  [1, 2, 5],  [2, 6],  [1, 1, 6]]这一题与之前的问题1来说， 只是不能重复使用数，所以在递归的时候i+1,而且因为有相同的数，为了规避这种情况引发的多个相同的解，要再遇到同样的数的时候跳过。
class Solution {    public List<List<Integer>> combinationSum2(int[] candidates, int target) {                 Arrays.sort(candidates);          List<List<Integer>> result = new ArrayList();          backtrack(result,new ArrayList<Integer>(),candidates,target,0);          return result;      }      public void backtrack(List<List<Integer>> result,List<Integer> list,int[] candidates,int target,int start){          if(target == 0){                      result.add(new ArrayList(list));                  }              if(target > 0){                  for(int i = start;i < candidates.length && target >= candidates[i];i++){                      if (i > start && candidates[i] == candidates[i - 1]){//如果是相同的数就可能引发同样的结果，所以将这种情况跳过                        continue;                    }                    list.add(candidates[i]);                      backtrack(result,list,candidates,target- candidates[i],i+1);                      list.remove(list.size()-1);                  }              }                                                }      }