From 81cee08f4bfcb87a0fea3b3b8ef4cc7d39c3369c Mon Sep 17 00:00:00 2001
From: chishouwu <u0151033@nkfust.edu.tw>
Date: Thu, 26 Aug 2021 21:12:22 +0800
Subject: [PATCH] add ago day 39 solution

---
 2021-08/2021-08-26 (day39)/AGO/README.md   | 22 ++++++++++++++++++
 2021-08/2021-08-26 (day39)/AGO/solution.ts | 27 ++++++++++++++++++++++
 2 files changed, 49 insertions(+)
 create mode 100644 2021-08/2021-08-26 (day39)/AGO/README.md
 create mode 100644 2021-08/2021-08-26 (day39)/AGO/solution.ts

diff --git a/2021-08/2021-08-26 (day39)/AGO/README.md b/2021-08/2021-08-26 (day39)/AGO/README.md
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+++ b/2021-08/2021-08-26 (day39)/AGO/README.md	
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+# 70. Climbing Stairs
+
+`Easy`
+
+## Description
+
+<p>You are climbing a staircase. It takes <code>n</code> steps to reach the top.</p>
+
+<p>Each time you can either climb <code>1</code> or <code>2</code> steps. In how many distinct ways can you climb to the top?</p>
+
+## My thought
+
+steps to 5 equals `step 3 combinations + 2 steps `or `step 3 combiations + 1 step`.
+
+So we can create a DP `f(n) = f(n-1) +　f(n-2)`
+There is two way
+
+1. Recursive from the end.
+2. Store steps spend into an array and loop from step 3 to end, plus previous steps and more previous steps.
+    `To save more menmory, we can just store 2 steps previous without whole route`
+
+## O(N)
diff --git a/2021-08/2021-08-26 (day39)/AGO/solution.ts b/2021-08/2021-08-26 (day39)/AGO/solution.ts
new file mode 100644
index 00000000..446ac4e7
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+++ b/2021-08/2021-08-26 (day39)/AGO/solution.ts	
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+// solution1
+function climbStairs(n: number): number {
+    let stairs:number[] = new Array();
+    stairs[1] = 1;
+    stairs[2] = 2;
+    
+    for(let i = 3; i<=n;i++){
+        stairs[i] = stairs[i-1] + stairs[i-2];
+    }
+    return stairs[n];
+};
+// solution2
+function climbStairs(n: number): number {
+    if (n==1) return 1;
+    if (n==2) return 2;
+    
+    let oneStepBefore = 2;
+    let twoStepBefore = 1;
+    let ans = 2;
+    for(let i = 3; i<=n;i++){
+        ans = oneStepBefore + twoStepBefore;
+        let temp = oneStepBefore;
+        oneStepBefore = ans;
+        twoStepBefore = temp;
+    }
+    return ans;
+};
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