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add q70/q104

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+69

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`@@ -69,6 +69,7 @@`
`69`	`69`
`70`	`70`	`* [q21_合并两个有序链表](/src/递归/q21_合并两个有序链表)`
`71`	`71`	`* [q101_对称二叉树](/src/递归/q101_对称二叉树)`
	`72`	`+* [q104_二叉树的最大深度](/src/递归/q104_二叉树的最大深度)`
`72`	`73`	`* [q226_翻转二叉树](/src/递归/q226_翻转二叉树)`
`73`	`74`	`* [q236_二叉树的最近公共祖先](/src/递归/q236_二叉树的最近公共祖先)`
`74`	`75`
`@@ -81,6 +82,7 @@`
`81`	`82`
`82`	`83`	`* [q5_最长回文子串](/src/动态规划/q5_最长回文子串)`
`83`	`84`	`* [q53_最大子序和](/src/动态规划/q53_最大子序和)`
	`85`	`+* [q70_爬楼梯](/src/动态规划/q70_爬楼梯)`
`84`	`86`	`* [q118_杨辉三角](/src/动态规划/q118_杨辉三角)`
`85`	`87`	`* [q300_最长上升子序列](/src/动态规划/q300_最长上升子序列)`
`86`	`88`	`* [q746_使用最小花费爬楼梯](/src/动态规划/q746_使用最小花费爬楼梯)`

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	`1`	`+package 动态规划.q70_爬楼梯;`
	`2`	`+`
	`3`	`+/**`
	`4`	`+ * 动态规划 dp[i]表示到达第i阶的方法总数dp[i]=dp[i−1]+dp[i−2] o(n)`
	`5`	`+ */`
	`6`	`+public class Solution {`
	`7`	`+`
	`8`	`+ public int climbStairs(int n) {`
	`9`	`+ if (n == 1) {`
	`10`	`+ return 1;`
	`11`	`+ }`
	`12`	`+ int[] dp = new int[n + 1];`
	`13`	`+ dp[1] = 1;`
	`14`	`+ dp[2] = 2;`
	`15`	`+ for (int i = 3; i <= n; i++) {`
	`16`	`+ dp[i] = dp[i - 1] + dp[i - 2];`
	`17`	`+ }`
	`18`	`+ return dp[n];`
	`19`	`+ }`
	`20`	`+}`

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	`1`	`+package 递归.q104_二叉树的最大深度;`
	`2`	`+`
	`3`	`+/**`
	`4`	`+ * 递归 o(n)`
	`5`	`+ */`
	`6`	`+public class Solution {`
	`7`	`+`
	`8`	`+ public int maxDepth(TreeNode root) {`
	`9`	`+ if (root == null) {`
	`10`	`+ return 0;`
	`11`	`+ } else {`
	`12`	`+ int leftHeight = maxDepth(root.left);`
	`13`	`+ int rightHeight = maxDepth(root.right);`
	`14`	`+ return Math.max(leftHeight, rightHeight) + 1;`
	`15`	`+ }`
	`16`	`+ }`
	`17`	`+}`

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