From cc0c8782d9a519292f8e3658135f298d3e71be7e Mon Sep 17 00:00:00 2001
From: zhicheng lee <904688436@qq.com>
Date: Tue, 6 Sep 2022 20:27:20 +0800
Subject: [PATCH] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200070.=E7=88=AC=E6=A5=BC?=
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三种Java代码中有两种是相同的方法，合并相同方法的内容
---
 problems/0070.爬楼梯.md | 37 ++++++++++++++-----------------------
 1 file changed, 14 insertions(+), 23 deletions(-)

diff --git a/problems/0070.爬楼梯.md b/problems/0070.爬楼梯.md
index 30c3642f..0a6acf7f 100644
--- a/problems/0070.爬楼梯.md
+++ b/problems/0070.爬楼梯.md
@@ -213,22 +213,6 @@ public:
 
 
 ### Java 
-```Java
-class Solution {
-    public int climbStairs(int n) {
-        // 跟斐波那契数列一样
-        if(n <= 2) return n;
-        int a = 1, b = 2, sum = 0;
-        
-        for(int i = 3; i <= n; i++){
-            sum = a + b;
-            a = b;
-            b = sum;
-        }
-        return b;
-    }
-}
-```
 
 ```java
 // 常规方式
@@ -241,15 +225,22 @@ public int climbStairs(int n) {
     }
     return dp[n];
 }
+```
+
+```Java
 // 用变量记录代替数组
-public int climbStairs(int n) {
-    int a = 0, b = 1, c = 0; // 默认需要1次
-    for (int i = 1; i <= n; i++) {
-        c = a + b;          // f(i - 1) + f(n - 2)
-        a = b;              // 记录上一轮的值
-        b = c;              // 向后步进1个数
+class Solution {
+    public int climbStairs(int n) {
+        if(n <= 2) return n;
+        int a = 1, b = 2, sum = 0;
+        
+        for(int i = 3; i <= n; i++){
+            sum = a + b;  // f(i - 1) + f(i - 2)
+            a = b;        // 记录f(i - 1)，即下一轮的f(i - 2)
+            b = sum;      // 记录f(i)，即下一轮的f(i - 1)
+        }
+        return b;
     }
-    return c;
 }
 ```