leetcode day3

最新推荐文章于 2025-02-10 16:59:43 发布

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本文介绍了两个经典算法问题：爬楼梯问题与路径求和问题的解决方案。爬楼梯问题通过动态规划算法计算到达楼顶的不同方式数量；路径求和问题则判断二叉树中是否存在从根节点到叶子节点的路径，使得路径上所有数值之和等于给定值。文章提供了简洁的代码实现。

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);

}