diff --git a/top-interview-leetcode150/dynamic-planning/198打家劫舍.js b/top-interview-leetcode150/dynamic-planning/198打家劫舍.js
new file mode 100644
index 0000000..5e97194
--- /dev/null
+++ b/top-interview-leetcode150/dynamic-planning/198打家劫舍.js
@@ -0,0 +1,38 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+const rob = function (nums) {
+
+};
+
+/*
+设dp[i]为偷取第i家时获取的最大利润，那么dp[i]可以由两种情况得来
+1.偷第i家，如果偷第i家，那么最大利润为 dp[i] = dp[i-2] + 第i家拥有的金额
+2.不偷第i家，如果不偷第i家，那么最大利润为 dp[i-1]的最大利润
+那么只需比较上面两种情况谁大，就是dp[i]的值，即dp[i] = Max(dp[i-1], dp[i-2] + nums[i])
+*/
+function f1(nums) {
+  const n = nums.length;
+  const dp = Array(n);
+  dp[0] = nums[0];
+  dp[1] = Math.max(nums[0], nums[1]);
+  for (let i = 2; i < dp.length; i++) {
+    dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);
+  }
+  return dp[n - 1];
+}
+
+/*
+使用双指针优化
+*/
+function f2(nums) {
+  let prev = 0;
+  let curr = 0;
+  for (const num of nums) {
+    const temp = Math.max(curr, prev + num);
+    prev = curr;
+    curr = temp;
+  }
+  return curr;
+}
diff --git a/top-interview-leetcode150/dynamic-planning/300最长递增子序列.js b/top-interview-leetcode150/dynamic-planning/300最长递增子序列.js
new file mode 100644
index 0000000..eb88c9b
--- /dev/null
+++ b/top-interview-leetcode150/dynamic-planning/300最长递增子序列.js
@@ -0,0 +1,30 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+const lengthOfLIS = function (nums) {
+
+};
+
+/*
+设dp[i]为以nums[i]结尾的递增子序列的最长长度,那么dp[i]，nums[i]可以接在nums[0-i)中比它小的元素
+组成子序列，求那个结果最长的就是dp[i],即dp[i] = Max(nums[0-i)) +1
+*/
+function f1(nums) {
+  const n = nums.length;
+  const dp = Array(n).fill(1);
+  dp[0] = 1;
+  for (let i = 1; i < n; i++) {
+    for (let j = 0; j < i; j++) {
+      if (nums[j] < nums[i]) {
+        // 更新dp[i]为最长子序列
+        dp[i] = Math.max(dp[j] + 1, dp[i]);
+      }
+    }
+  }
+  console.log(dp);
+
+  return dp[n - 1];
+}
+
+f1([1, 3, 6, 7, 9, 4, 10, 5, 6]);
diff --git a/top-interview-leetcode150/dynamic-planning/70爬楼梯.js b/top-interview-leetcode150/dynamic-planning/70爬楼梯.js
new file mode 100644
index 0000000..ecb05e1
--- /dev/null
+++ b/top-interview-leetcode150/dynamic-planning/70爬楼梯.js
@@ -0,0 +1,41 @@
+/**
+ * @param {number} n
+ * @return {number}
+ */
+const climbStairs = function (n) {
+
+};
+
+/*
+这个题目就是求斐波那契数列的现实抽象，我们的最终目的是跳到第n个台阶，而每一次要么跳一步，要么跳两步，
+所以智能由n-1个台阶跳一步上来，或者由n-2个台阶跳两步上来，设dp[i]为跳转到这个台阶可能的方法，那么
+dp[i] = dp[i-1] + dp[i-2]
+*/
+function f1(n) {
+  // 定义dp数组
+  const dp = Array(n + 1);
+  dp[1] = 1;
+  dp[2] = 2;
+  for (let i = 3; i <= n; i++) {
+    dp[i] = dp[i - 1] + dp[i - 2];
+  }
+  return dp[n];
+}
+
+/*
+通过观察上面的代码可以知道，dp[i]的结果只由它的前一个位置和前前个位置有关，所以定义两个变量first和second来
+保存这两个值
+*/
+function f2(n) {
+  if (n < 3) return n;
+  // 定义跳到前两个台阶可能的步数
+  let first = 1;
+  let second = 2;
+  let result = 1;
+  for (let i = 3; i <= n; i++) {
+    result = first + second;
+    first = second;
+    second = result;
+  }
+  return result;
+}