LeetCode(40)组合总和Ⅱ??

给定一个候选人编号的集合?candidates?和一个目标数?target?，找出?candidates?中所有可以使数字和为?target?的组合。

candidates?中的每个数字在每个组合中只能使用?一次?。

注意：解集不能包含重复的组合。?

示例?1:

输入: candidates =?[10,1,2,7,6,1,5], target =?8,
输出:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]

示例?2:

输入: candidates =?[2,5,2,1,2], target =?5,
输出:
[
[1,2,2],
[5]
]

提示:

• 1 <=?candidates.length <= 100
• 1 <=?candidates[i] <= 50
• 1 <= target <= 30

思路：和39一样的思路，注意回溯

class Solution {
          List<List<Integer>> result = new ArrayList<>();
          LinkedList<Integer> path = new LinkedList<>();
          public List<List<Integer>> combinationSum2(int[] candidates, int target) {
              if (candidates == null || candidates.length == 0){
                  return result;
              }
              Arrays.sort(candidates);
              backTrachking(candidates,target,0);
              return result;
          }
          public void backTrachking(int[] candidates,int target,int index){
              //终止条件
              if (target == 0){
                  result.add(new ArrayList<>(path));
                  return;
              }
              if (target < 0){
                  return;
              }
              if (index == candidates.length){
                  return;
              }
              for (int i = index; i < candidates.length; i++) {
                  //正确剔除重复解的办法
                  //跳过同一树层使用过的元素
                  if ( i > index && candidates[i] == candidates[i - 1] ) {
                      continue;
                  }
                  int curNum = candidates[i];
                  if (target - curNum < 0){
                      break;
                  }
                  path.add(curNum);
                  backTrachking(candidates,target - curNum,i + 1);//这里加一是不能重复使用
                  path.removeLast();
              }
          }
}