找到最长的子字符串而不重复字符

时间:2015-02-12 20:18:51

标签: c++ arrays string algorithm

我写了这个程序:

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int i = 0, t_max = 0, max = 0;
        char arr[256];
        fill_n(arr, 256, -1);

        for (i = 0; i < s.length(); i++) {
            if (arr[(int)s[i]] != -1) {
                if (i != s.length() - 1){
                    t_max = t_max >= (i - arr[(int)s[i]]) ? i - arr[(int)s[i]] : ++t_max;
                    arr[(int)s[i]] = i;
                } else {
                    t_max = t_max >= (i - arr[(int)s[i]]) ? t_max : ++t_max; 
                }
            } else {
                arr[(int)s[i]] = i;
                t_max++;
            }
            max = max > t_max ? max : t_max;
        }
        return max;
    }
};

for leetcode

除了以下所有测试用例的通过:(如果您有任何想法,请告诉我):

"abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~ 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3 个答案:

答案 0 :(得分:2)

那个测试用例很长。很长。像20,000个字符的东西。您的算法依赖于在数组中存储字符串索引,但您的数组定义为:

char arr[256];

那根本无法容纳那么大的价值观。只需将其添加到:

int arr[256];

应该处理它。

那就是说,你的代码存在问题。这两行都类似:

t_max = t_max >= (i - arr[(int)s[i]]) ? t_max : ++t_max; 

是未定义的行为,因为左侧的t_max读取与右侧++t_max的副作用无关。见sequence points。这可以很容易地替换为t_max + 1

此外,您的所有索引都是通过arr[(int)s[i]]完成的。你不需要那里演员。只需arr[s[i]]即可。

答案 1 :(得分:2)

我认为你让事情过于复杂,出了什么问题:

public:
int lengthOfLongestSubstring(string s) {
    int i = 0, max = 0, offset = -1;
    int arr[256];
    fill_n(arr, 256, -1);
    for (i = 0; i < s.length(); i++) {
        int si = (int) s[i];
        offset = std::max(offset,arr[si]);
        max = std::max(max,i-offset);
        arr[si] = i;
    } 
    return max;
}

offset存储子字符串开始的位置。对于每个角色,我们会查找角色最后一次出现的时间。如果它比最后一个偏移更新,则将偏移设置为该出现。

此外max存储当前最大值。该最大值始终从offset到当前索引计算。我们还将最后一次出现在数组中。在循环结束时,返回最大值。

答案 2 :(得分:0)

我发现大多数答案都很难理解。 这是我的c ++解决方案。它有效,但我不确定时间和空间的复杂程度,如果有人可以评论它。

##标题##

   class Solution {
     public:
      int lengthOfLongestSubstring(string s) {
      int count=0;
      for (int i=0; i < s.length(); i++) {
          unordered_map <char, int> hashmaptest;
          int  j=0;
          while(hashmaptest.find(s[i+j]) == hashmaptest.end() 
                && j< s.length()-i) {
            hashmaptest[s[i+j]]=i+j;
            j++;
           }

          count=max(count, int(hashmaptest.size()) );    
          }


         return count;    
        }



          };