CombinationSum

Change-Id: Iec3a3b218ec60ba68dbc07bf18709c2e77d06266

Author: applewjg

CombinationSumII.java

@@ -0,0 +1,45 @@
+/*
+ Author:     King, [email protected]
+ Date:       Dec 20, 2014
+ Problem:    Combination Sum II
+ Difficulty: Easy
+ Source:     https://oj.leetcode.com/problems/combination-sum-ii/
+ Notes:
+ 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] 
+
+ Solution: ..Similar to Combination Sum I, except for line 42 && 44.
+ */
+public class Solution {
+    public List<List<Integer>> combinationSum2(int[] num, int target) {
+        List<List<Integer>> res = new ArrayList<List<Integer>>();
+        Arrays.sort(num);
+        ArrayList<Integer> path = new ArrayList<Integer>();
+        combinationSumRe(num, target, 0, path, res);
+        return res;
+    }
+    void combinationSumRe(int[] candidates, int target, int start, ArrayList<Integer> path, List<List<Integer>> res) {
+        if (target == 0) {
+            ArrayList<Integer> p = new ArrayList<Integer>(path);
+            res.add(p);
+            return;
+        }
+        for (int i = start; i < candidates.length && target >= candidates[i]; ++i) {
+            if (i!=start && candidates[i-1] == candidates[i]) continue;
+            path.add(candidates[i]);
+            combinationSumRe(candidates, target-candidates[i], i+1, path, res);
+            path.remove(path.size() - 1);
+        }
+    }
+}