leetcode day3

70. Climbing Stairs

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Note: Given n will be a positive integer.

Example 1:

Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

 

talk is cheap let me show you the code

 

public int climbStairs(int n) {         if (n <=2) {             return n;         } else {             return climbStairs(n-1) + climbStairs(n-2);         }   }

 

结果 

 

更简的写法，动态规划

talk is cheap let me show you the code

 

public int climbStairs(int n) {           if (n == 1) {            return 1;        }        int []dp = new int[n+1];        dp[1] =1;        dp[2] =2;        for (int i=3; i<=n; i++) {            dp[i] = dp[i-1] + dp[i-2];        }        return dp[n];    }

 

112. Path Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

 

 

talk is cheap, let me show you the code

 

public boolean hasPathSum(TreeNode root, int sum) {     if (root == null ) {         return false;     }     sum = sum - root.val;     if (root.left == null && root.right == null) {         return sum == 0;     }      return  hasPathSum(root.left,sum) || hasPathSum(root.right,sum);   }