查找不包含重复字符的最长子字符串

Find the longest substring without repeating characters

本文关键字：字符串 字符 包含重 查找      更新时间：2023-10-16

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;
    }
};

用于leetcode

它通过了所有的测试用例，除了这个：（如果你有任何想法，请告诉我）：

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那个测试用例很长。很长。大约20000个字符。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];

这根本不能保持那么大的价值观。简单地将其提升为：

int arr[256];

也就是说，您的代码有问题。两条类似的线路：

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CCD_ 1的读取相对于右侧的++t_max的副作用是未排序的。请参见序列点。这可以很容易地替换为t_max + 1。

此外，所有索引都是通过arr[(int)s[i]]完成的。你不需要那个石膏。只要arr[s[i]]就可以了。

我认为你把事情搞得太复杂了，怎么了

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存储子字符串的起始位置。对于每个角色，我们查找该角色最后出现的时间。如果它发生的时间比上一次偏移更近，则偏移设置为该次发生。

此外，max存储当前最大值。该最大值总是从offset到当前索引来计算的。我们还将最后一次出现的情况存储在数组中。在循环结束时，返回最大值。

我发现大多数答案都很难理解。这是我的c++解决方案。它是有效的，但我不确定时间和空间的复杂性，如果有人可以对此发表评论的话

##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;    
        }

          };
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