题目: For a given source string and a target string, you should output the first index(from 0) of target string in source string. If target does not exist in source, just return -1. Clarification Do I need to implement KMP Algorithm in a real interview…
Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. 思路: 注意,在for循环中条件有计算得到的负数时, 一定要把计算括起来转换为int, 否则会默认转换为uchar 负数就会被误认为是一个很大的数字. ; i < - ); ++i) 实现很常规: int strStr(string haystac…
Implement strStr() Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. 最简单的思路,逐一比较: class Solution { public: int strStr(char *haystack, char *needle) { int n1=strlen(haystack);…
Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. 解题: 本题为典型的KMP算法考察题,KMP算法描述为: 设主串S,匹配串P,i为S的索引下标,j为P的索引下标,初始i和j设置为0. 在i小于S的长度和j小于P的长度时,开始循环: 1.比较S[i]和P[j]是否相同: 2.如果相同则i++,j+…
题目: The code base version is an integer start from 1 to n. One day, someone committed a bad version in the code case, so it caused this version and the following versions are all failed in the unit tests. Find the first bad version. You can call isBa…
问题描述: 比较两个字符串A和B,确定A中是否包含B中所有的字符.字符串A和B中的字符都是 大写字母. 样例 给出 A = "ABCD" B = "ACD",返回 true 给出 A = "ABCD" B = "AABC", 返回 false 注意事项 在 A 中出现的 B 字符串里的字符不需要连续或者有序. 问题分析: 实质上利用的是哈希表的思想.只有大写字母,一共26个,遍历A的时候,往里面压,遍历B的时候,往外边弹,如果…