90 add benchmark test

Author: aQuaYi

Algorithms/0090.subsets-ii/README.md

@@ -1,14 +1,13 @@
 # [90. Subsets II](https://leetcode.com/problems/subsets-ii/)
 
-## 题目
 Given a collection of integers that might contain duplicates, nums, return all possible subsets.
 
 Note: The solution set must not contain duplicate subsets.
 
-
 For example,
-If nums = [1,2,2], a solution is:
 
+```text
+If nums = [1,2,2], a solution is:
 
 [
   [2],
@@ -18,8 +17,4 @@ If nums = [1,2,2], a solution is:
   [1,2],
   []
 ]
-
-## 解题思路
-观察题目输出结果的规律后，利用dfs解题
-
-见程序注释
+```

Algorithms/0090.subsets-ii/subsets-ii.go

@@ -2,27 +2,25 @@ package problem0090
 
 import "sort"
 
-func subsetsWithDup(nums []int) [][]int {
+func subsetsWithDup(A []int) [][]int {
 	res := [][]int{}
-	sort.Ints(nums)
+	sort.Ints(A)
 
 	var dfs func(int, []int)
-	dfs = func(idx int, temp []int) {
+	dfs = func(index int, temp []int) {
 		t := make([]int, len(temp))
 		copy(t, temp)
-		// 没有以上两行，答案就是错的
-		// 因为temp的底层数组在递归过程中，不停地修改
-		// 程序结束时，temp的底层数组的值，全部是 nums 的最大值。
 		res = append(res, t)
 
-		for i := idx; i < len(nums); i++ {
-			if i == idx || nums[i] != nums[i-1] {
-				dfs(i+1, append(temp, nums[i]))
+		for i := index; i < len(A); i++ {
+			if i == index ||
+				A[i] != A[i-1] { // 避免重复答案
+				dfs(i+1, append(temp, A[i]))
 			}
 		}
 	}
 
-	temp := make([]int, 0, len(nums))
+	temp := make([]int, 0, len(A))
 	dfs(0, temp)
 
 	return res

Algorithms/0090.subsets-ii/subsets-ii_test.go

@@ -74,3 +74,10 @@ func Test_Problem0090(t *testing.T) {
 		ast.Equal(a.one, subsetsWithDup(p.nums), "输入:%v", p)
 	}
 }
+
+func Benchmark_subsetsWithDup(b *testing.B) {
+	A := []int{1, 2, 3, 6, 5, 4, 7, 8, 9, 5, 6, 3, 2, 4}
+	for i := 1; i < b.N; i++ {
+		subsetsWithDup(A)
+	}
+}