diff --git a/Algorithms/1049. Last Stone Weight II/1049. Last Stone Weight II.go b/Algorithms/1049. Last Stone Weight II/1049. Last Stone Weight II.go
new file mode 100644
index 00000000..c93a17b0
--- /dev/null
+++ b/Algorithms/1049. Last Stone Weight II/1049. Last Stone Weight II.go	
@@ -0,0 +1,22 @@
+package leetcode
+
+func lastStoneWeightII(stones []int) int {
+	sum := 0
+	for _, v := range stones {
+		sum += v
+	}
+	n, C, dp := len(stones), sum/2, make([]int, sum/2+1)
+	for i := 0; i <= C; i++ {
+		if stones[0] <= i {
+			dp[i] = stones[0]
+		} else {
+			dp[i] = 0
+		}
+	}
+	for i := 1; i < n; i++ {
+		for j := C; j >= stones[i]; j-- {
+			dp[j] = max(dp[j], dp[j-stones[i]]+stones[i])
+		}
+	}
+	return sum - 2*dp[C]
+}
diff --git a/Algorithms/1049. Last Stone Weight II/1049. Last Stone Weight II_test.go b/Algorithms/1049. Last Stone Weight II/1049. Last Stone Weight II_test.go
new file mode 100644
index 00000000..0084f922
--- /dev/null
+++ b/Algorithms/1049. Last Stone Weight II/1049. Last Stone Weight II_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1049 struct {
+	para1049
+	ans1049
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1049 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1049 struct {
+	one int
+}
+
+func Test_Problem1049(t *testing.T) {
+
+	qs := []question1049{
+
+		question1049{
+			para1049{[]int{2, 7, 4, 1, 8, 1}},
+			ans1049{1},
+		},
+
+		question1049{
+			para1049{[]int{21, 26, 31, 33, 40}},
+			ans1049{5},
+		},
+
+		question1049{
+			para1049{[]int{1, 2}},
+			ans1049{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1049------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1049, q.para1049
+		fmt.Printf("【input】:%v       【output】:%v\n", p, lastStoneWeightII(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/1049. Last Stone Weight II/README.md b/Algorithms/1049. Last Stone Weight II/README.md
new file mode 100755
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+++ b/Algorithms/1049. Last Stone Weight II/README.md	
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+# [1049. Last Stone Weight II](https://leetcode.com/problems/last-stone-weight-ii/)
+
+## 题目:
+
+We have a collection of rocks, each rock has a positive integer weight.
+
+Each turn, we choose **any two rocks** and smash them together. Suppose the stones have weights `x` and `y` with `x <= y`. The result of this smash is:
+
+- If `x == y`, both stones are totally destroyed;
+- If `x != y`, the stone of weight `x` is totally destroyed, and the stone of weight `y`has new weight `y-x`.
+
+At the end, there is at most 1 stone left. Return the **smallest possible** weight of this stone (the weight is 0 if there are no stones left.)
+
+**Example 1:**
+
+    Input: [2,7,4,1,8,1]
+    Output: 1
+    Explanation: 
+    We can combine 2 and 4 to get 2 so the array converts to [2,7,1,8,1] then,
+    we can combine 7 and 8 to get 1 so the array converts to [2,1,1,1] then,
+    we can combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
+    we can combine 1 and 1 to get 0 so the array converts to [1] then that's the optimal value.
+
+**Note:**
+
+1. `1 <= stones.length <= 30`
+2. `1 <= stones[i] <= 100`
+
+
+
+## 题目大意
+
+有一堆石头，每块石头的重量都是正整数。每一回合，从中选出任意两块石头，然后将它们一起粉碎。假设石头的重量分别为 x 和 y，且 x <= y。那么粉碎的可能结果如下：
+
+如果 x == y，那么两块石头都会被完全粉碎；
+如果 x != y，那么重量为 x 的石头将会完全粉碎，而重量为 y 的石头新重量为 y-x。
+最后，最多只会剩下一块石头。返回此石头最小的可能重量。如果没有石头剩下，就返回 0。
+
+提示：
+
+1. 1 <= stones.length <= 30
+2. 1 <= stones[i] <= 1000
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面的元素代表的是石头的重量。现在要求两个石头对碰，如果重量相同，两个石头都消失，如果一个重一个轻，剩下的石头是两者的差值。问经过这样的多次碰撞以后，能剩下的石头的重量最轻是多少？
+- 由于两两石头要发生碰撞，所以可以将整个数组可以分为两部分，如果这两部分的石头重量总和相差不大，那么经过若干次碰撞以后，剩下的石头重量一定是最小的。现在就需要找到这样两堆总重量差不多的两堆石头。这个问题就可以转化为 01 背包问题。从数组中找到 `sum/2` 重量的石头集合，如果一半能尽量达到 `sum/2`，那么另外一半和 `sum/2` 的差是最小的，最好的情况就是两堆石头的重量都是 `sum/2`，那么两两石头对碰以后最后都能消失。01 背包的经典模板可以参考第 416 题。