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+# [70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs/description/)
+# 思路
+简单的动态规划。   
+按照最后是跨一步还是两步可以将到达第n步的所有情况分为两种：
+* 1、从第n-2步阶梯一下跨两步到第n步阶梯；
+* 2、从第n-1步阶梯跨一步到第n步阶梯；    
+即若dp[i]代表跨到第i步阶梯的情况数，那么`dp[i] = dp[i - 1] + dp[i - 2]`。  
+时间复杂度和空间复杂度都为O(n)，可将空间复杂度优化为O(1)
+# C++
+```
+class Solution {
+public:
+    int climbStairs(int n) {
+        vector<int>dp(n + 1);
+        dp[0] = 1;
+        dp[1] = 1;
+        for(int i = 2; i <= n;i++){
+            dp[i] = dp[i - 1] + dp[i - 2];
+        }
+        return dp[n];
+    }
+};
+```
+空间优化后的版本:
+```
+class Solution {
+public:
+    int climbStairs(int n) {
+        int tmp, pre = 1, res = 1;
+        for(int i = 2; i <= n;i++){
+            tmp = res;
+            res += pre;
+            pre = tmp;
+        }
+        return res;
+    }
+};
+```