From be33e5b3c92f9c3d228ff15402a7c6913d599337 Mon Sep 17 00:00:00 2001
From: nanhuaibeian <49868746+nanhuaibeian@users.noreply.github.com>
Date: Wed, 12 May 2021 20:59:49 +0800
Subject: [PATCH] =?UTF-8?q?Update=200040.=E7=BB=84=E5=90=88=E6=80=BB?=
 =?UTF-8?q?=E5=92=8CII.md?=
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---
 problems/0040.组合总和II.md | 36 +++++++++++++++++++++++++++++++++
 1 file changed, 36 insertions(+)

diff --git a/problems/0040.组合总和II.md b/problems/0040.组合总和II.md
index ffcbe212..50898016 100644
--- a/problems/0040.组合总和II.md
+++ b/problems/0040.组合总和II.md
@@ -255,7 +255,43 @@ public:
 
 
 Java：
+```Java
+class Solution {
+    List<List<Integer>> lists = new ArrayList<>();
+    Deque<Integer> deque = new LinkedList<>();
+    int sum = 0;
 
+    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
+        //为了将重复的数字都放到一起，所以先进行排序
+        Arrays.sort(candidates);
+        //加标志数组，用来辅助判断同层节点是否已经遍历
+        boolean[] flag = new boolean[candidates.length];
+        backTracking(candidates, target, 0, flag);
+        return lists;
+    }
+
+    public void backTracking(int[] arr, int target, int index, boolean[] flag) {
+        if (sum == target) {
+            lists.add(new ArrayList(deque));
+            return;
+        }
+        for (int i = index; i < arr.length && arr[i] + sum <= target; i++) {
+            //出现重复节点，同层的第一个节点已经被访问过，所以直接跳过
+            if (i > 0 && arr[i] == arr[i - 1] && !flag[i - 1]) {
+                continue;
+            }
+            flag[i] = true;
+            sum += arr[i];
+            deque.push(arr[i]);
+            //每个节点仅能选择一次，所以从下一位开始
+            backTracking(arr, target, i + 1, flag);
+            int temp = deque.pop();
+            flag[i] = false;
+            sum -= temp;
+        }
+    }
+}
+```
 
 Python：