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CombinationSum.java
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59 lines (48 loc) · 1.77 KB
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import java.util.ArrayList;
/*
Given a set of candidate numbers (C) and a target number (T),
find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
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 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]
*/
public class CombinationSum {
public static void main(String[] args) {
int[] a = new int[]{2, 3, 5, 6, 7, 10};
int target = 10;
ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
ArrayList<Integer> r = new ArrayList<Integer>();
combinationSum(a, 0, target, r, results);
System.out.println(results.toString());
}
public static void combinationSum(int[] a,
int index,
int target,
ArrayList<Integer> r,
ArrayList<ArrayList<Integer>> results) {
// find it: target == 0
if (target == 0) {
results.add(new ArrayList<Integer>(r));
return;
}
if (a.length == index)
return;
combinationSum(a, index + 1, target, r, results);
int tmp = a[index];
int canUse = target / tmp;
for (int i = 1; i <= canUse; ++i) {
r.add(tmp);
combinationSum(a, index + 1, target - tmp * i, r, results);
}
for (int i = 1; i <= canUse; ++i) {
r.remove(Integer.valueOf(tmp));
}
}
}