Add solution 070

Author: yanglbme

README.md

@@ -16,6 +16,7 @@ Complete solutions to Leetcode problems, updated daily.
 | 007 | [Reverse Integer](https://github.com/yanglbme/leetcode/tree/master/solution/007.Reverse%20Integer) | `Math` |
 | 013 | [Roman to Integer](https://github.com/yanglbme/leetcode/tree/master/solution/013.Roman%20to%20Integer) | `Math`, `String` |
 | 021 | [Merge Two Sorted Lists](https://github.com/yanglbme/leetcode/tree/master/solution/021.Merge%20Two%20Sorted%20Lists) | `Linked List` |
+| 070 | [Climbing Stairs](https://github.com/yanglbme/leetcode/tree/master/solution/070.Climbing%20Stairs) | `Dynamic Programming` |
 | 083 | [Remove Duplicates from Sorted List](https://github.com/yanglbme/leetcode/tree/master/solution/083.Remove%20Duplicates%20from%20Sorted%20List) | `Linked List` |
 | 198 | [House Robber](https://github.com/yanglbme/leetcode/tree/master/solution/198.House%20Robber) | `Dynamic Programming` |
 | 203 | [Remove Linked List Elements](https://github.com/yanglbme/leetcode/tree/master/solution/203.Remove%20Linked%20List%20Elements) | `Linked List` |

solution/70.Climbing Stairs/README.md

@@ -0,0 +1,47 @@
+## 爬楼梯
+### 题目描述
+
+假设你正在爬楼梯。需要 n 阶你才能到达楼顶。
+
+每次你可以爬 1 或 2 个台阶。你有多少种不同的方法可以爬到楼顶呢？
+
+注意：给定 n 是一个正整数。
+
+示例 1：
+```
+输入： 2
+输出： 2
+解释： 有两种方法可以爬到楼顶。
+1.  1 阶 + 1 阶
+2.  2 阶
+```
+
+示例 2：
+```
+输入： 3
+输出： 3
+解释： 有三种方法可以爬到楼顶。
+1.  1 阶 + 1 阶 + 1 阶
+2.  1 阶 + 2 阶
+3.  2 阶 + 1 阶
+```
+
+### 解法
+爬上 1 阶有 1 种方法，爬上 2 阶有 2 种方法，爬上 n 阶 f(n) 有 f(n - 1) + f(n - 2) 种方法。可以利用数组记录中间结果，防止重复计算。
+
+```java
+class Solution {
+    public int climbStairs(int n) {
+        if (n < 3) {
+            return n;
+        }
+        int[] res = new int[n + 1];
+        res[1] = 1;
+        res[2] = 2;
+        for (int i = 3; i < n + 1; ++i) {
+            res[i] = res[i - 1] + res[i - 2];
+        }
+        return res[n];
+    }
+}
+```

solution/70.Climbing Stairs/Solution.java

@@ -0,0 +1,14 @@
+class Solution {
+    public int climbStairs(int n) {
+        if (n < 3) {
+            return n;
+        }
+        int[] res = new int[n + 1];
+        res[1] = 1;
+        res[2] = 2;
+        for (int i = 3; i < n + 1; ++i) {
+            res[i] = res[i - 1] + res[i - 2];
+        }
+        return res[n];
+    }
+}