添加 0040.组合总和II.md 中的Java与Go的不使用标记数组的解法

problems/0040.组合总和II.md

@@ -255,6 +255,7 @@ public:
 
 
 ## Java 
+**使用标记数组**
 ```Java
 class Solution {
     List<List<Integer>> lists = new ArrayList<>();
@@ -292,6 +293,44 @@ class Solution {
     }
 }
 ```
+**不使用标记数组**
+```Java
+class Solution {
+  List<List<Integer>> res = new ArrayList<>();
+  LinkedList<Integer> path = 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 );
+    backTracking( candidates, target, 0 );
+    return res;
+  }
+  
+  private void backTracking( int[] candidates, int target, int start ) {
+    if ( sum == target ) {
+      res.add( new ArrayList<>( path ) );
+      return;
+    }
+    for ( int i = start; i < candidates.length && sum + candidates[i] <= target; i++ ) {
+      //正确剔除重复解的办法
+      //跳过同一树层使用过的元素
+      if ( i > start && candidates[i] == candidates[i - 1] ) {
+        continue;
+      }
+
+      sum += candidates[i];
+      path.add( candidates[i] );
+      // i+1 代表当前组内元素只选取一次
+      backTracking( candidates, target, i + 1 );
+
+      int temp = path.getLast();
+      sum -= temp;
+      path.removeLast();
+    }
+  }
+}
+```
 
 ## Python 
 **回溯+巧妙去重(省去使用used**
@@ -384,6 +423,7 @@ class Solution:
 ## Go
 主要在于如何在回溯中去重
 
+**使用used数组**
 ```go
 func combinationSum2(candidates []int, target int) [][]int {
     var trcak []int
@@ -423,7 +463,41 @@ func backtracking(startIndex,sum,target int,candidates,trcak []int,res *[][]int,
     }
 }
 ```
-
+**不使用used数组**
+```go
+func combinationSum2(candidates []int, target int) [][]int {
+    var trcak []int
+    var res [][]int
+    sort.Ints(candidates)
+    backtracking(0,0,target,candidates,trcak,&res)
+    return res
+}
+func backtracking(startIndex,sum,target int,candidates,trcak []int,res *[][]int){
+    //终止条件
+    if sum==target{
+        tmp:=make([]int,len(trcak))
+        //拷贝
+        copy(tmp,trcak)
+        //放入结果集
+        *res=append(*res,tmp)
+        return
+    }
+    //回溯
+    for i:=startIndex;i<len(candidates) && sum+candidates[i]<=target;i++{
+        // 若当前树层有使用过相同的元素，则跳过
+        if i>startIndex&&candidates[i]==candidates[i-1]{
+                continue
+        }
+        //更新路径集合和sum
+        trcak=append(trcak,candidates[i])
+        sum+=candidates[i]
+        backtracking(i+1,sum,target,candidates,trcak,res)
+        //回溯
+        trcak=trcak[:len(trcak)-1]
+        sum-=candidates[i]
+    }
+}
+```
 ## javaScript
 
 ```js