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请教：要办I-485，那么I-693的体检在休斯顿哪儿能做？
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请教：要办I-485，那么I-693的体检在休斯顿哪儿能做？# Immigration - 落地生根
m*1
1 楼
这个题最后为了避免重复，我用了一个set结构，有什么更好的方法来避免重复吗？
avatar
e*m
2 楼
请教：要办I-485，那么I-693的体检在休斯顿哪儿能做？
有没有网站能查到？多谢！
avatar
p*2
3 楼
这题可能要求不可以用extra space。
avatar
K*M
4 楼
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avatar
m*1
5 楼
那循环已有的查查看有无重复？

【在 p*****2 的大作中提到】
: 这题可能要求不可以用extra space。
avatar
e*m
6 楼
Thanks
avatar
c*r
7 楼
请教二爷，不用extra space如何避免呢？

【在 p*****2 的大作中提到】
: 这题可能要求不可以用extra space。
avatar
e*m
8 楼
请问：是不是只需做皮试、x-ray，还有打三针疫苗？
不会好多项目吧？大约多少钱？多谢！
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p*2
9 楼

要不要我现写一个呀。

【在 c********r 的大作中提到】
: 请教二爷，不用extra space如何避免呢？
avatar
e*o
10 楼
I think it is similar to combination sum I.
sort the array first.
in combination sum I, you can use a number unlimited times.
but in II, if a number appears n times, then you can use it at most n times
.
avatar
p*2
11 楼

times
确实similar。代码非常类似。

【在 e******o 的大作中提到】
: I think it is similar to combination sum I.
: sort the array first.
: in combination sum I, you can use a number unlimited times.
: but in II, if a number appears n times, then you can use it at most n times
: .
avatar
s*s
12 楼
同问如何不用extra hashset避免duplicate。
avatar
r*e
13 楼
my code which passed online judge:
void solveSum2(vector &num, int tgt, vector > &res, int sum
, int idx, vector &index) {
if (sum>tgt) return ;
if (sum==tgt) {
vector one;
for (size_t k=1; kone.push_back(num[index[k]]);
res.push_back(one);
return;
}
for (size_t i=index[idx]; iif (sum>0 && i==(size_t)index[idx]) continue;//avoid adding the same
beginning index one more time
index.push_back(i);
solveSum2(num, tgt, res, sum+num[i], idx+1, index);
int back=num[index.back()];
while (i+1index.pop_back();
}
}
vector > combSum2(vector &num, int tgt) {
vector > res;
vector index;
index.push_back(0);//need this!
sort(num.begin(), num.end());
solveSum2(num, tgt, res, 0, 0, index);
return res;
}

【在 m*******1 的大作中提到】
: 这个题最后为了避免重复，我用了一个set结构，有什么更好的方法来避免重复吗？
avatar
m*1
14 楼
这个题最后为了避免重复，我用了一个set结构，有什么更好的方法来避免重复吗？
avatar
p*2
15 楼
这题可能要求不可以用extra space。
avatar
m*1
16 楼
那循环已有的查查看有无重复？

【在 p*****2 的大作中提到】
: 这题可能要求不可以用extra space。
avatar
c*r
17 楼
请教二爷，不用extra space如何避免呢？

【在 p*****2 的大作中提到】
: 这题可能要求不可以用extra space。
avatar
p*2
18 楼

要不要我现写一个呀。

【在 c********r 的大作中提到】
: 请教二爷，不用extra space如何避免呢？
avatar
e*o
19 楼
I think it is similar to combination sum I.
sort the array first.
in combination sum I, you can use a number unlimited times.
but in II, if a number appears n times, then you can use it at most n times
.
avatar
p*2
20 楼

times
确实similar。代码非常类似。

【在 e******o 的大作中提到】
: I think it is similar to combination sum I.
: sort the array first.
: in combination sum I, you can use a number unlimited times.
: but in II, if a number appears n times, then you can use it at most n times
: .
avatar
s*s
21 楼
同问如何不用extra hashset避免duplicate。
avatar
r*e
22 楼
my code which passed online judge:
void solveSum2(vector &num, int tgt, vector > &res, int sum
, int idx, vector &index) {
if (sum>tgt) return ;
if (sum==tgt) {
vector one;
for (size_t k=1; kone.push_back(num[index[k]]);
res.push_back(one);
return;
}
for (size_t i=index[idx]; iif (sum>0 && i==(size_t)index[idx]) continue;//avoid adding the same
beginning index one more time
index.push_back(i);
solveSum2(num, tgt, res, sum+num[i], idx+1, index);
int back=num[index.back()];
while (i+1index.pop_back();
}
}
vector > combSum2(vector &num, int tgt) {
vector > res;
vector index;
index.push_back(0);//need this!
sort(num.begin(), num.end());
solveSum2(num, tgt, res, 0, 0, index);
return res;
}

【在 m*******1 的大作中提到】
: 这个题最后为了避免重复，我用了一个set结构，有什么更好的方法来避免重复吗？
avatar
t*e
23 楼
有没有java solution么？下面的代码是combination sum的，怎么在这个基础上改呢？
public class Solution {
public ArrayList> combinationSum(int[] candidates,
int target) {
// Start typing your Java solution below
// DO NOT write main() function
ArrayList> results = new ArrayListInteger>>();
Arrays.sort(candidates);
addUp(candidates, 0, target, new ArrayList(), results);
return results;
}
private void addUp(int[] candidates, int start, int target, ArrayList<
Integer> path, ArrayList> results) {
if (start < 0 || start >= candidates.length || target < 0) return;
if (target == 0) {
ArrayList res = new ArrayList(path);
results.add(res);
} else {
for (int i=start; iif (candidates[i] > target) continue; // we cannot break
since data may not be sorted
path.add(candidates[i]);
addUp(candidates, i, target - candidates[i], path, results);
path.remove(path.size() - 1);
}
}
}
}
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