leetcode-40-Combination-Sum-II

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描述

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:

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

分析

这道题和 leetcode-39-Combination-Sum基本是一样的，只是这一题要求 candidates 中的内容不能重复出现，只需要修改两处代码。

解决方案1(C++)

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class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>> result;
        vector<int> path;
        dfs(candidates, path, result, target, 0);
        return result;
    }
    
private:
    void dfs(vector<int>& candidates, vector<int>& path, vector<vector<int>>& result, int target, int start) {
        if(target == 0) {
            result.push_back(path);
            return;
        }
        
        for(int i = start; i < candidates.size(); i++) {
            if(i>start && candidates[i] == candidates[i-1]) {
                continue;
            }
            if(candidates[i] > target) {
                return;
            }
            path.push_back(candidates[i]);
            dfs(candidates, path, result, target-candidates[i], i+1);
            path.pop_back();
        }
    }
};

解决方案2(Python)

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class Solution(object):
    def combinationSum2(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        candidates.sort()
        result = []
        path = []
        self.dfs(candidates, path, result, target)
        return result
        
    def dfs(self, candidates, path, result, target):
        if target == 0:
            result.append(path[:])
            return
        pre = None 
        for item, value in enumerate(candidates):
            if pre is None or pre != value:
                if target < value:
                    return 
                path.append(value)
                self.dfs(candidates[item+1:], path, result, target-value)
                path.pop()
            pre = value

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