BAT面試算法進階(2)- 無重複字符的最長子串(暴力法)

一.算法題

• 題目

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.算法題解讀

• 題目大意:給定一個字符串,找出不含有重複字符的最長子串的長度

• 解讀Example

• 給定"abcabcbb",沒有重複字符的最長子串是"abc",那麼長度就是3

• 給定"bbbbb",最長子串就是"b",長度就是1
• 給定pwwkew,最長子串就是"wke",長度爲3,
• ==注意,==必須是一個子串."pwke",是子序列,而不是子串

三.暴力解決方案

3.1 思路

逐個檢查全部的子字符串,看它是否不含有重複字符

3.2 算法

爲了枚舉給定字符串的全部子字符串,咱們須要枚舉它們開始和結束的索引,假如開始和結束的索引分別是i和j.那麼咱們有0<=i<=j<=n.所以,使用i從0到n-1以及j從i+1到n這2個嵌套循環.咱們就能夠遍歷出a的全部子字符串.

3.3 複雜的分析

• 時間複雜度:o(n3);
• 空間複雜度:o(min(n,m));

3.4 參考代碼

//(2)無重複字符的最長子串
//求字符串長度函數
int strLength(char *p)
{
    int number = 0;
    while (*p) {
        
        number++;
        p++;
    }
    return number;
    
}

//判斷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)無重複子串的最長子串
    char *s = "pwwkew";
    int n = LengthLongestSubstring(s);
    printf("%d",n);
    
    return 0;
}
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