LeetCode40. Combination Sum II

一、题目

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sum to target.

Each number in candidates may only be used once in the combination.

Note: The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8
Output:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
Example 2:

Input: candidates = [2,5,2,1,2], target = 5
Output:
[
[1,2,2],
[5]
]

Constraints:

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

二、题解

使用used数组进行去重

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    void backtracking(vector<int>& candidates,int target,int sum,int startIndex,vector<int>& used){
        if(sum > target) return;
        if(sum == target){
            res.push_back(path);
            return;
        }
        for(int i = startIndex;i < candidates.size();i++){
            if(i > 0 && candidates[i-1] == candidates[i] && used[i-1] == 0)continue;
            sum += candidates[i];
            used[i] = 1;
            path.push_back(candidates[i]);
            backtracking(candidates,target,sum,i+1,used);
            sum -= candidates[i];
            used[i] = 0;
            path.pop_back();
        }
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        int n = candidates.size();
        vector<int> used(n,0);
        sort(candidates.begin(),candidates.end());
        backtracking(candidates,target,0,0,used);
        return res;
    }
};

使用startIndex进行去重

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