40. Combination Sum II -Medium

Question

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.

给出一个候选数字集合C和目标值T，请你找出所有C中元素相加得到T的组合。（每个元素只能使用一次，且C中有相同元素）

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]
]

Solution

• 回溯解。因为元素不能多次使用，所以在回溯过程中，每次从下个元素的索引开始，而又因为C中有相同元素，所以我们需要先对数组进行排序，然后每当遇到第二个相同元素作为起始点时则跳过不做计算

  public class Solution {
      public List> combinationSum2(int[] candidates, int target) {
          Arrays.sort(candidates);
          List> res = new ArrayList<>();
          backtracking(candidates, 0, target, res, new ArrayList<>());
          return res;
      }
  
      public void backtracking(int[] candidates, int start, int target, List> res, List temp){
          if(target < 0) return;
          if(target == 0){
              res.add(new ArrayList<>(temp));
              return;
          }
          for(int i = start; i < candidates.length; i++){
              // 如果出现重复元素，第二个元素直接跳过
              if(i > start && candidates[i] == candidates[i - 1]) continue;
              // 仅当当前元素小于目标值时选择
              if(candidates[i] <= target) {
                  temp.add(candidates[i]);
                  // 每次指向下一个元素的索引
                  backtracking(candidates, i + 1, target - candidates[i], res, temp);
                  temp.remove(temp.size() - 1);
              }
          }
      }
  }