[LeetCode] 2. Add Two Numbers 两个数字相加Grandyang刷尽天下 立即播放 打开App,流畅又高清100+个相关视频 更多157 -- 9:59 App [LeetCode] 1. Two Sum 两数之和 126 -- 9:53 App [LeetCode] 82. Remove Duplicates from Sorted List II 删除排序链表中的重复元素之二 149 -- 20:30 App [...
类似题目:LeetCode 67 - Add Binary | 二进制求和 (Rust) 时间复杂度:O(|l1| + |l2|) 需要遍历 l1 中的全部 O(|l1|) 个结点 需要遍历 l2 中的全部 O(|l2|) 个结点 空间复杂度:O(1) 需要为结果链表中的全部 O(max(|l1|, |l2|)) 个结点分配空间 (理论上可以复用已有的结点,这样就只需要定...
445. Add Two Numbers II 这个题和Add Two Numbers相似,但是这个题是把最高位放在了链表的前面,但数字的相加必须从最低位开始。 可以选择用栈来存储数字,也可以使用链表反转。 使用链表反转,先将l1、l2反转,然后时候用和Add Two Numbers一样的方法做计算,然后再反转这个生成的记过就好了。 classSolution {public...
Github 同步地址: https://github.com/grandyang/leetcode/issues/67 类似题目: Add Binary Multiply Strings Plus One Linked List Plus One Add Two Numbers Add to Array-Form of Integer 参考资料: https://leetcode.com/problems/add-binary/ https://leetcode.com/problems/add-binary/discuss/24475/short...
Leetcode 之Add Binary(29) 比较简单,细节:先将字符串翻转,注意进位。 stringaddBinary(stringa,stringb) {stringresult;intlen = a.size() > b.size() ?a.size() : b.size(); reverse(a.begin(), a.end()); reverse(b.begin(), b.end());intcarry =0;for(inti =0; i < len; i++)...
Question: You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) ...
1111-Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings 1118-Number-of-Days-in-a-Month 1119-Remove-Vowels-from-a-String 1120-Maximum-Average-Subtree 1121-Divide-Array-Into-Increasing-Sequences 1124-Longest-Well-Performing-Interval .gitignore qrcode.png readme.md Breadcrumbs ...
【Leetcode】Power of Two 题目: Given an integer, write a function to determine if it is a power of two. 思路: 如果小于0,则不是2的n次幂。 否则则转为二进制,只有最高位为1,剩余为0的时候才是2的n次幂。 算法: AI检测代码解析 public boolean isPowerOfTwo(int n) {...
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Leetcode - Add Binary Paste_Image.png My code: public class Solution { public String addBinary(String a, String b) { if (a == null || a.length() == 0 || b == null || b.length() == 0) return null; int len = Math.max(a.length(), b.length()) + 1;...