一.算法题
- 题目
Given a string, find the length of the longest substring without repeating characters.
- Example
- Given "abcabcbb", the answer is "abc", which the length is 3.
- Given "bbbbb", the answer is "b", with the length of 1.
- Given "pwwkew", the answer is "wke", with the length of
- Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
二.算法题解读
-
题目大意:给定一个字符串,找出不含有重复字符的最长子串的长度
- 解读Example
- 给定"abcabcbb",没有重复字符的最长子串是"abc",那么长度就是3
- 给定"bbbbb",最长子串就是"b",长度就是1
- 给定pwwkew,最长子串就是"wke",长度为3,
- ==注意,==必须是一个子串."pwke",是子序列,而不是子串
三.暴力解决方案
3.1 思路
逐个检查所有的子字符串,看它是否不含有重复字符
3.2 算法
为了枚举给定字符串的所有子字符串,我们需要枚举它们开始和结束的索引,假如开始和结束的索引分别是i和j.那么我们有0<=i<=j<=n.因此,使用i从0到n-1以及j从i+1到n这2个嵌套循环.我们就可以遍历出a的所有子字符串.
3.3 复杂的分析
- 时间复杂度:
o(n3); - 空间复杂度:
o(min(n,m));
3.4 参考代码
//(2)无重复字符的最长子串
//求字符串长度函数
int strLength(char *p)
{
int number = 0;
while (*p) {
number++;
p++;
}
return number;
}
//判断子字符在字符串中是否唯一
int unRepeatStr(char *a,int start,int end)
{
for (int i=start;i<end xss=removed xss=removed xss=removed xss=removed xss=removed xss=removed xss=removed xss=removed> j-i)?ans:j-i;
}
}
}
return ans;
}
int main(int argc, const char * argv[]) {
//2)无重复子串的最长子串
char *s = "pwwkew";
int n = LengthLongestSubstring(s);
printf("%d",n);
return 0;
}小编给大家推荐一个iOS技术交流群：551346706！群内提供数据结构与算法、底层进阶、swift、逆向、底层面试题整合文档等免费资料！