From 121028b95585231a8c6fd077b0cbe6062008a934 Mon Sep 17 00:00:00 2001
From: whaifree <49432110+whaibetter@users.noreply.github.com>
Date: Tue, 30 Apr 2024 12:55:51 +0800
Subject: [PATCH] =?UTF-8?q?=E4=BF=AE=E5=A4=8D=E4=BA=86LeetCode300=E5=92=8C?=
 =?UTF-8?q?LeetCode115=E7=9A=84=E6=B5=8B=E8=AF=95=E4=BB=A3=E7=A0=81?=
 =?UTF-8?q?=EF=BC=8C=E4=BC=98=E5=8C=96=E4=BA=86=E6=8F=90=E4=BA=A4=E4=BF=A1?=
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---
 .../redo/redo_24_4_27/LeetCode115.java        | 65 +++++++++++++++++++
 .../redo/redo_24_4_27/LeetCode300.java        | 52 +++++++++++++++
 2 files changed, 117 insertions(+)
 create mode 100644 src/main/java/cn/whaifree/redo/redo_24_4_27/LeetCode115.java
 create mode 100644 src/main/java/cn/whaifree/redo/redo_24_4_27/LeetCode300.java

diff --git a/src/main/java/cn/whaifree/redo/redo_24_4_27/LeetCode115.java b/src/main/java/cn/whaifree/redo/redo_24_4_27/LeetCode115.java
new file mode 100644
index 0000000..86450a7
--- /dev/null
+++ b/src/main/java/cn/whaifree/redo/redo_24_4_27/LeetCode115.java
@@ -0,0 +1,65 @@
+package cn.whaifree.redo.redo_24_4_27;
+
+import org.junit.Test;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/30 11:20
+ * @注释
+ */
+public class LeetCode115 {
+    @Test
+    public void test() {
+        String s = "babgbag";
+        String t = "bag";
+        System.out.println(new Solution().numDistinct(s, t));
+    }
+
+
+
+    class Solution {
+        /**
+         * 给你两个字符串 s 和 t ，统计并返回在 s 的 子序列 中 t 出现的个数，结果需要对 109 + 7 取模。
+         * @param s
+         * @param t
+         * @return
+         */
+        public int numDistinct(String s, String t) {
+            /**
+             *
+             * dp[i][j] 表示0-j在0-i子序列中出现的次数
+             *    '' b a b g b a g
+             *''  1  1 1 1 1 1 1 1
+             * b  0  1 1 2 2 3 3 3
+             * a  0  0 1 1 1 1 4 4
+             * g  0  0 0 0 1 1 1 5
+             *
+             * if s[i] == t[j]
+             *      dp[i][j] =
+             *           +
+             *           - s使用i-1个位置 dp[i-1][j-1]
+             *           - s不使用i-1个位置 dp[i][j-1]
+             * else
+             *      dp[i][j-1]
+             */
+
+            int lenS = s.length();
+            int lenT = t.length();
+            int[][] dp = new int[lenT + 1][lenS + 1];
+            for (int i = 0; i <= lenS; i++) {
+                dp[0][i] = 1;
+            }
+            for (int j = 1; j <= lenT; j++) {
+                for (int i = 1; i <= lenS; i++) {
+                    if (s.charAt(i - 1) == t.charAt(j - 1)) {
+                        dp[j][i] = dp[j - 1][i - 1] + dp[j][i - 1];
+                    } else {
+                        dp[j][i] = dp[j][i - 1];
+                    }
+                }
+            }
+            return dp[lenT][lenS];
+        }
+    }
+}
diff --git a/src/main/java/cn/whaifree/redo/redo_24_4_27/LeetCode300.java b/src/main/java/cn/whaifree/redo/redo_24_4_27/LeetCode300.java
new file mode 100644
index 0000000..776373b
--- /dev/null
+++ b/src/main/java/cn/whaifree/redo/redo_24_4_27/LeetCode300.java
@@ -0,0 +1,52 @@
+package cn.whaifree.redo.redo_24_4_27;
+
+import org.junit.Test;
+
+import java.util.Arrays;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/30 11:49
+ * @注释
+ */
+public class LeetCode300 {
+    @Test
+    public void test()
+    {
+        Solution solution = new Solution();
+        int[] nums = new int[]{10, 9, 2, 5, 3, 7, 101, 18};
+        int res = solution.lengthOfLIS(nums);
+        System.out.println(res);
+    }
+
+    class Solution {
+        /**
+         * 最长严格递增子序列的长度。
+         *
+         * dp[i]表示以i为结尾的最长递增子序列的长度 （计算i的时候，0~(i-1)已经满足最长递增子序列的长度）
+         *
+         *
+         * if (nums[j] < nums[i])
+         *      dp[i] = dp[j] + 1
+         *      dp[j] = dp[i]
+         *
+         * @param nums
+         * @return
+         */
+        public int lengthOfLIS(int[] nums) {
+            int[] dp = new int[nums.length];
+            Arrays.fill(dp, 1);
+            int res = 1;
+            for (int i = 1; i < nums.length; i++) {
+                for (int j = 0; j < i; j++) {
+                    if (nums[j] < nums[i]) {
+                        dp[i] = Math.max(dp[i], dp[j] + 1);
+                    }
+                }
+                res = Math.max(res, dp[i]);
+            }
+            return res;
+        }
+    }
+}