add q40

Author: yuanguangxin

README.md

@@ -101,6 +101,7 @@
 
 * [q10_正则表达式匹配](/src/回溯法/q10_正则表达式匹配)
 * [q22_括号生成](/src/回溯法/q22_括号生成)
+* [q40_组合总和2](/src/回溯法/q40_组合总和2)
 * [q46_全排列](/src/回溯法/q46_全排列)
 
 ### 树的遍历

src/回溯法/q40_组合总和2/Solution.java

@@ -0,0 +1,38 @@
+package 回溯法.q40_组合总和2;
+
+import java.util.*;
+
+/**
+ * 回溯法 O(n*log(n))
+ */
+class Solution {
+
+    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
+        List<List<Integer>> res = new ArrayList<>();
+        if (candidates.length == 0) {
+            return res;
+        }
+        Arrays.sort(candidates);
+        helper(candidates, target, 0, new LinkedList<>(), res);
+        return res;
+    }
+
+    public void helper(int[] candidates, int target, int start, LinkedList<Integer> stack, List<List<Integer>> res) {
+        if (start > candidates.length) {
+            return;
+        }
+        if (target == 0 && !stack.isEmpty()) {
+            List<Integer> item = new ArrayList<>(stack);
+            res.add(item);
+        }
+        HashSet<Integer> set = new HashSet<>();
+        for (int i = start; i < candidates.length; ++i) {
+            if (!set.contains(candidates[i]) && target >= candidates[i]) {
+                stack.push(candidates[i]);
+                helper(candidates, target - candidates[i], i + 1, stack, res);
+                stack.pop();
+                set.add(candidates[i]);
+            }
+        }
+    }
+}