Leetcode-Combinations Sum II
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]
public class Solution {
public List<List<Integer>> combinationSum2(int[] num, int target) {
int[] candidates = num;
List<List<Integer>> resSet = new ArrayList<List<Integer>>();
List<Integer> curRes = new ArrayList<Integer>();
if (candidates.length==0) return resSet;
Arrays.sort(candidates);
int cur=0,end=candidates.length-1;
for (int i=0;i<candidates.length;i++)
if (candidates[i]>target){
end = i-1;
break;
} sumRecur(candidates,cur,end,target,resSet,curRes); return resSet; } public void sumRecur(int[] candidates, int cur, int end, int valLeft, List<List<Integer>> resSet, List<Integer> curRes){
if (valLeft==0){
List<Integer> temp = new ArrayList<Integer>();
temp.addAll(curRes);
resSet.add(temp);
return;
} if (cur>end) return; int newLeft = valLeft;
int curLen = curRes.size();
int nextIndex = cur;
while (nextIndex<=end && candidates[nextIndex]==candidates[cur]) nextIndex++; for (int i=cur;i<nextIndex;i++)
if (newLeft>=candidates[i]){
curRes.add(candidates[i]);
newLeft -= candidates[i];
sumRecur(candidates,nextIndex,end,newLeft,resSet,curRes);
} else
break; while (curRes.size()!=curLen) curRes.remove(curRes.size()-1); sumRecur(candidates,nextIndex,end,valLeft,resSet,curRes);
}
}
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