<p>Given a collection of candidate numbers (<code>candidates</code>) and a target number (<code>target</code>), find all unique combinations in <code>candidates</code>&nbsp;where the candidate numbers sum to <code>target</code>.</p>

<p>Each number in <code>candidates</code>&nbsp;may only be used <strong>once</strong> in the combination.</p>

<p><strong>Note:</strong>&nbsp;The solution set must not contain duplicate combinations.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> candidates = [10,1,2,7,6,1,5], target = 8
<strong>Output:</strong> 
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> candidates = [2,5,2,1,2], target = 5
<strong>Output:</strong> 
[
[1,2,2],
[5]
]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;=&nbsp;candidates.length &lt;= 100</code></li>
	<li><code>1 &lt;=&nbsp;candidates[i] &lt;= 50</code></li>
	<li><code>1 &lt;= target &lt;= 30</code></li>
</ul>

class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        ans = []
        path = []
        n = len(candidates)
        candidates.sort()
        self.cur_sum = 0

        def dfs(i):
            if self.cur_sum == target:
                ans.append(path.copy())
                return

            if self.cur_sum > target: #没有这个会time limit exceed
                return 

            for j in range(i, n):
                if j> i and candidates[j]==candidates[j-1]:
                    continue
                path.append(candidates[j])
                self.cur_sum += candidates[j]
                dfs(j+1)
                path.pop()
                self.cur_sum -= candidates[j]
        dfs(0)
        return ans 

        
16700