leetcode-algorithms-3 Longest Substring Without Repeating Characters
leetcode-algorithms-3 Longest Substring Without Repeating Characters
Given a string, find the length of the longest substring without repeating characters.
Example 1:
Input: "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Example 2:
Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:
Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
解法1
一步步检查所有字串看是否有重复的字符
class Solution
{
public:
int lengthOfLongestSubstring(string s)
{
string longest;
for (int i = 0; i < s.size(); ++i)
{
for (int j = s.size() - i; j > 0; --j)
{
string longesttemp = s.substr(i, j);
bool repect = false;
int a[256] = {0};
for (int n = 0; n < longesttemp.size(); ++n)
{
int x = longesttemp[n];
++a[x];
if (a[x] > 1)
{
repect = true;
break;
}
}
if (!repect && (longesttemp.size() > longest.size())) longest = longesttemp;
}
}
return longest.size();
}
};
时间复杂度: O(n^3).
空间复杂度: O(min(n,m)).
解法2
解法1的时间复杂度太高了,效率低下.字符串查找要怎么提高效率,首先都应该想到Sliding Window算法.设定一个窗口[i, j],将i到j的内容存入map,下面滑动j,如果s[j] (s表示字符串)在map中存在,表示字符串已经重复了,记下最大值,然后将i窗口滑到s[j]上次的位置.
下面是代码的实现:
class Solution
{
public:
int lengthOfLongestSubstring(string s)
{
int n = s.length();
std::unordered_map<char, int> m;
int len = 0;
for (int i = 0, j = 0; j < n; ++j)
{
auto fiter = m.find(s[j]);
if (fiter != m.end())
{
i = (fiter->second) > i ? fiter->second : i;
}
int temp_len = j - i + 1;
len = (len > temp_len) ? len : temp_len;
m[s[j]] = j + 1;
}
return len;
}
};
时间复杂度: O(n).只有一个j循环.
空间复杂度: O(m).m是最大的不重复子串的长度.
解法3
对于解法2有没更快捷的方式.参考KMP算法对字符串的处理,我们可以将字符存入整型数组,值存储字符所在的位置,这样就可以在算法2的基础上少一次查询.
class Solution
{
public:
int lengthOfLongestSubstring(string s)
{
int n = s.length();
int len = 0;
int index[128] = {0};
for (int i = 0, j = 0; j < n; ++j)
{
i = i > index[s[j]] ? i : index[s[j]];
int temp_len = j - i + 1;
len = (len > temp_len) ? len : temp_len;
index[s[j]] = j + 1;
}
return len;
}
};
时间复杂度: O(n).
空间复杂度: O(128).
