Combination Sum
Given asetof candidate numbers (candidates
)(without duplicates)and a target number (target
), find all unique combinations incandidates
where the candidate numbers sums totarget
.
Thesamerepeated number may be chosen fromcandidates
unlimited number of times.
Note:
All numbers (including
target
) will be positive integers.The solution set must not contain duplicate combinations.
Example 1:
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]
Example 2:
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
Analysis
Solution
Backtracking
这里技巧在于每次candidates的搜索空间要减小一些,但是又因为candidates[]元素可以重复利用,因此进入下一层递归时,用startIdx = i,而不是 startIdx = i + 1。
class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> ans = new ArrayList<>();
Arrays.sort(candidates);
helper(target, 0, 0, candidates, new ArrayList<>(), ans);
return ans;
}
private void helper(int target, int sum, int startIdx, int[] candidates,
List<Integer> combo, List<List<Integer>> ans) {
if (sum == target) {
ans.add(new ArrayList<Integer>(combo));
return;
}
if (sum > target) {
return;
}
for (int i = startIdx; i < candidates.length; i++) {
combo.add(candidates[i]);
// the startIdx is not i + 1 since we can reuse same elements
helper(target, sum + candidates[i], i, candidates, combo, ans);
combo.remove(combo.size() - 1);
}
}
}
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