From 165fd427339ced15bc7b4e812d5c18f52638cb2f Mon Sep 17 00:00:00 2001
From: Guanzhong Pan <gpan20020205@gmail.com>
Date: Wed, 12 Jan 2022 15:59:27 +0000
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 =?UTF-8?q?=E6=A2=AF.md=20C=E8=AF=AD=E8=A8=80=E7=89=88=E6=9C=AC?=
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---
 problems/0070.爬楼梯.md | 42 ++++++++++++++++++++++++++++++++++++++
 1 file changed, 42 insertions(+)

diff --git a/problems/0070.爬楼梯.md b/problems/0070.爬楼梯.md
index 97db670a..7f47a991 100644
--- a/problems/0070.爬楼梯.md
+++ b/problems/0070.爬楼梯.md
@@ -296,6 +296,48 @@ var climbStairs = function(n) {
 };
 ```
 
+### C
+```c
+int climbStairs(int n){
+    //若n<=2，返回n
+    if(n <= 2)
+        return n;
+    //初始化dp数组，数组大小为n+1
+    int *dp = (int *)malloc(sizeof(int) * (n + 1));
+    dp[0] = 0, dp[1] = 1, dp[2] = 2;
+
+    //从前向后遍历数组，dp[i] = dp[i-1] + dp[i-2]
+    int i;
+    for(i = 3; i <= n; ++i) {
+        dp[i] = dp[i - 1] + dp[i - 2];
+    }
+    //返回dp[n]
+    return dp[n];
+}
+```
+
+优化空间复杂度：
+```c
+int climbStairs(int n){
+    //若n<=2，返回n
+    if(n <= 2)
+        return n;
+    //初始化dp数组，数组大小为3
+    int *dp = (int *)malloc(sizeof(int) * 3);
+    dp[1] = 1, dp[2] = 2;
+
+    //只记录前面两个台阶的状态
+    int i;
+    for(i = 3; i <= n; ++i) {
+        int sum = dp[1] + dp[2];
+        dp[1] = dp[2];
+        dp[2] = sum;
+    }
+    //返回dp[2]
+    return dp[2];
+}
+```
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>