40. Combination Sum II

发表于 | 分类于

每日一题 2019 - 03 - 27

题目：

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

Each number in candidates 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.

Example 1:

1
2
3
4
5
6
7
8
Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:

1
2
3
4
5
6
Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

解法：

这个题让我们在给定数组里面找到不重复的序列的和等于给定的值，跟昨天的sum I不同的地方在于这个题元素不可重复，那个题是可重复的，所以思路还是回溯，不过为了保证输出元素的有序性，可以先把 candidates 中得元素先 sort 一下，还有一个需要注意的地方就是可能res中存放的元素有重复的，所以就在push到res进去之前，先遍历一遍res看是否存在重复的序列，或者使用set

代码：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
class Solution {
public:
    vector<vector<int>>res;
    vector<int>ans;
    int nums_len;

    vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
        sort(candidates.begin(),candidates.end());
        if(candidates.empty()) return res;
        nums_len = candidates.size();
        dfs(0, 0, target, candidates);
        return res;
    }

    void dfs(int start, int sum, int target,vector<int> candidates){
        if(sum == target) {
            if(res.end() != find(res.begin(),res.end(),ans))
            {
                return ;
            }
            res.push_back(ans); return; }
        else if(sum > target)
            return;
        else{
            for(int i = start; i < nums_len ; i++){
                ans.push_back(candidates[i]);
                dfs(i + 1, sum + candidates[i], target, candidates);
                ans.pop_back();
            }
        }
    }
};
0%
undefined