diff --git a/src/main/java/leetcode/editor/cn/NumberOfLongestIncreasingSubsequence.java b/src/main/java/leetcode/editor/cn/NumberOfLongestIncreasingSubsequence.java
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+//给定一个未排序的整数数组，找到最长递增子序列的个数。 
+//
+// 示例 1: 
+//
+// 
+//输入: [1,3,5,4,7]
+//输出: 2
+//解释: 有两个最长递增子序列，分别是 [1, 3, 4, 7] 和[1, 3, 5, 7]。
+// 
+//
+// 示例 2: 
+//
+// 
+//输入: [2,2,2,2,2]
+//输出: 5
+//解释: 最长递增子序列的长度是1，并且存在5个子序列的长度为1，因此输出5。
+// 
+//
+// 注意: 给定的数组长度不超过 2000 并且结果一定是32位有符号整数。 
+// Related Topics 树状数组 线段树 数组 动态规划 👍 352 👎 0
+
+package leetcode.editor.cn;
+
+import java.util.Arrays;
+
+//673:最长递增子序列的个数
+class NumberOfLongestIncreasingSubsequence {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new NumberOfLongestIncreasingSubsequence().new Solution();
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+
+        public int findNumberOfLIS(int[] nums) {
+            int N = nums.length;
+            if (N <= 1) return N;
+            int[] lengths = new int[N];
+            int[] counts = new int[N];
+            Arrays.fill(counts, 1);
+            for (int j = 0; j < N; ++j) {
+                for (int i = 0; i < j; ++i) {
+                    if (nums[i] < nums[j]) {
+                        if (lengths[i] >= lengths[j]) {
+                            lengths[j] = lengths[i] + 1;
+                            counts[j] = counts[i];
+                        } else if (lengths[i] + 1 == lengths[j]) {
+                            counts[j] += counts[i];
+                        }
+                    }
+                }
+            }
+            int longest = 0, ans = 0;
+            for (int length : lengths) {
+                longest = Math.max(longest, length);
+            }
+            for (int i = 0; i < N; ++i) {
+                if (lengths[i] == longest) {
+                    ans += counts[i];
+                }
+            }
+            return ans;
+        }
+    }
+    //leetcode submit region end(Prohibit modification and deletion)
+
+}
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