BAT面试算法进阶(2)- 无重复字符的最长子串(暴力法)

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

逐个检查全部的子字符串,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全部子字符串,咱们须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.那么咱们有0<=i<=j<=n.所以,使用i从0到n-1以及j从i+1到n这2个嵌套循环.咱们就能够遍历出a的全部子字符串.

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惟一
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;
}
复制代码

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