LeetCode 40. Combination Sum II

原题链接在这里：https://leetcode.com/problems/combination-sum-ii/

题目：

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：

与Combination Sum非常相似 也是backtracking. 不同在于不可以重复使用元素。其实只是递归时, start的参数更改为i+1即可.

Time Complexity: exponential.

Space: O(candidates.length). stack space.

AC Java:

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if(candidates == null || candidates.length == 0){
            return res;
        }
        
        Arrays.sort(candidates);
        dfs(candidates, target, 0, new ArrayList<Integer>(), res);
        return res;
    }
    
    private void dfs(int[] candidates, int target, int start, List<Integer> item, List<List<Integer>> res){
        if(target == 0){
            res.add(new ArrayList<Integer>(item));
            return;
        }
        
        if(target < 0){
            return;
        }
        
        for(int i = start; i<candidates.length; i++){
            if(i>start && candidates[i]==candidates[i-1]){
                continue;
            }
            
            item.add(candidates[i]);
            dfs(candidates, target-candidates[i], i+1, item, res);
            item.remove(item.size()-1);
        }
    }
}

AC Python:

class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        res = []
        candidates.sort()
        self.dfs(candidates, target, 0, [], res)
        return res
    
    def dfs(self, candidates: List[int], target: int, start: int, item: List[int], res: List[List[int]]) -> None:
        if target == 0:
            res.append(item[:])
            return
        
        for i in range(start, len(candidates)):
            if candidates[i] > target:
                continue
                
            if i > start and candidates[i] == candidates[i - 1]:
                continue
                
            item.append(candidates[i])
            self.dfs(candidates, target - candidates[i], i + 1, item, res)
            item.pop()

跟上Combination Sum III.