【Combination Sum II 】cpp
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.
- Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
- 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]
代码:
class Solution {
public:
vector<vector<int> > combinationSum2(vector<int>& candidates, int target)
{
vector<vector<int> > ret;
std::sort(candidates.begin(), candidates.end());
int sum = ;
vector<int> tmp;
Solution::dfs(ret, tmp, sum, candidates, , candidates.size()-, target);
return ret;
}
static void dfs(
vector<vector<int> >& ret,
vector<int>& tmp,
int &sum,
vector<int>& candidates,
int begin,
int end,
int target
)
{
if ( sum>target ) return;
if ( sum==target )
{
ret.push_back(tmp);
return;
}
int pre = candidates[]-;
for ( int i=begin; i<=end; ++i )
{
if ( pre==candidates[i] ) continue;
pre = candidates[i];
if ( sum+candidates[i]<=target )
{
sum += candidates[i];
tmp.push_back(candidates[i]);
Solution::dfs(ret, tmp, sum, candidates, i+, end, target);
tmp.pop_back();
sum -= candidates[i];
}
}
}
};
tips:
此题与combination sum不同之处在于,每个元素只能取一次,并且解集中不能有重复的。
用深搜模板:
1. 如果元素都没有重复的,就是最直接的深搜模板(注意dfs到下一层的时候,传入的begin是i+1而不是i了)
2. 如果元素有重复的,再处理时就跳过前面出现过的元素(前提是candidates都排好序);这里有个技巧就是维护一个pre变量,并且初始化pre为candidates[0]-1,即比candidates元素都小,这样不用改变循环的结构就可以直接处理
3. 还有一个疑问,为什么不用判断begin>end的情况?比如,{1,1,1} ,4 这种输入,显然所有元素加一起也满足不了结果。进行到最后一定会出现begin==3 end==2的情况。这种情况也不要紧,因为begin大于end就不处理了,直接返回了,所以也没事。
===========================================
第二次过这道题,遇到这种不要重复结果的,有些规律:就是在每一层dfs的时候,如果candidate[i]出现连续两个重复,就跳过后面的那个。
class Solution {
public:
vector<vector<int> > combinationSum2(
vector<int>& candidates, int target)
{
sort(candidates.begin(), candidates.end());
vector<vector<int> > ret;
vector<int> tmp;
Solution::dfs(ret, tmp, candidates, , candidates.size()-, target);
return ret;
}
static void dfs(
vector<vector<int> >& ret,
vector<int>& tmp,
vector<int>& candidates,
int begin,
int end,
int target
)
{
if ( target< ) return;
if ( target== )
{
ret.push_back(tmp);
return;
}
if ( begin>end ) return;
int pre = candidates[begin]-;
for ( int i=begin; i<=end; ++i )
{ if ( pre==candidates[i]) continue;
pre = candidates[i];
tmp.push_back(candidates[i]);
Solution::dfs(ret, tmp, candidates, i+, end, target-candidates[i]);
tmp.pop_back();
}
}
};
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