All elements of each string will have an ASCII value in[97, 122]. 这道题给了我们两个字符串,让我们删除一些字符使得两个字符串相等,我们希望删除的字符的ASCII码最小。这道题跟之前那道Delete Operation for Two Strings极其类似,那道题让求删除的最少的字符数,这道题换成了ASCII码值。其实很多大厂的面...
1finalclassSolution {2func minimumDeleteSum(_ s1: String, _ s2: String) ->Int {3let s1 =Array(s1.utf8).map(Int.init)4let s2 =Array(s2.utf8).map(Int.init)5vardp = [[Int]](repeating: Array(repeating:0, count: s2.count &+1), count: s1.count &+1)67foriin0..<s1.count {...
Explanation: Deleting "dee" from "delete" to turn the string into "let", adds 100[d]+101[e]+101[e] to the sum. Deleting "e" from "leet" adds 101[e] to the sum. At the end, both strings are equal to "let", and the answer is 100+101+101+101 = 403. If instead we turned...
示例2: 输入:s1 = "delete", s2 = "leet"输出:403解释:在 "delete" 中删除 "dee" 字符串变成 "let", 将 100[d]+101[e]+101[e] 加入总和。在 "leet" 中删除 "e" 将 101[e] 加入总和。 结束时,两个字符串都等于 "let",结果即为 100+101+101+101 = 403 。 如果改为将两个字符串转换为...
LeetCode之371. Sum of Two Integers --- 使用位运算实现加法: a^b 加不同部分 (a&b)<<1 加相同部分 递归相加 AC代码: publicclassSolution {publicintgetSum(inta,intb) {if(b==0)returna;intt1=a^b;intt2=(a&b)<<1;returngetSum(t1,t2); } } 1. 2. 3. 4. 5. 6. 7. 8. 题目...
Input:s = "leetcode", k = 2 Output:6 Explanation: The operations are as follows: - Convert: "leetcode" ➝ "(12)(5)(5)(20)(3)(15)(4)(5)" ➝ "12552031545" ➝ 12552031545 - Transform #1: 12552031545 ➝ 1 + 2 + 5 + 5 + 2 + 0 + 3 + 1 + 5 + 4 + 5 ➝ ...
题目描述参见leetcode 解法1:扫描两个列表,如果发现第一个列表的字符串和第二个列表的相等,而且下标之和也小于等于之前记录的最小值,那么就把答案加入,这样子就有可能记录局部最小值的答案,所以需要记录每次再记录下当前答案的局部最小值,最后筛掉就可以了 {{{ class Solution { public: vector<string> findRest...
Input: s1 = “delete”, s2 = “leet” Output: 403 Explanation: Deleting “dee” from “delete” to turn the string into “let”, adds 100[d]+101[e]+101[e] to the sum. Deleting “e” from “leet” adds 101[e] to the sum. At the end, both strings are equal to “let”, ...
Memory Usage: 41.3 MB, less than 10.92% of Java online submissions for Two Sum. 优化解法 优化思路: 借助HashMap 数据结构,将原本 O(n) 的遍历,降低为 O(1) 的查询。 java 实现如下: 实现时注意 HashMap 的扩容问题,此处直接指定为和数组一样大。 public int[] twoSum(int[] nums, int target)...
You are given a string s consisting of digits and an integer k. A round can be completed if the length of s is greater than k. In one round, do the following: Divide s into consecutive groups of size k such that the first k characters are in the first group, the next k ...