table of Contents
Source of subject: https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/
Title description
Given a string, you find out which does not contain a repeated character longest substring length.
示例 1:
输入: s = "abcabcbb"
输出: 3
解释: 因为无重复字符的最长子串是 "abc"，所以其长度为 3。
示例 2:
输入: s = "bbbbb"
输出: 1
解释: 因为无重复字符的最长子串是 "b"，所以其长度为 1。
示例 3:
输入: s = "pwwkew"
输出: 3
解释: 因为无重复字符的最长子串是 "wke"，所以其长度为 3。
请注意，你的答案必须是 子串 的长度，"pwke" 是一个子序列，不是子串。
示例 4:
输入: s = ""
输出: 0
提示：
0 <= s.length <= 5 * 104
s 由英文字母、数字、符号和空格组成
General idea
- Find the longest string that does not contain repeated characters, use a cnt array that records the number of occurrences, through the form of sliding window + double pointer
Sliding window + dual pointer
- Use an array (or hash table) to count, maintain a variable-length window, and constantly move the left and right pointers to take the longest window length
const int MAXN = 200;
int cnt[MAXN];
class Solution {
public:
int lengthOfLongestSubstring(string s) {
int ans = 0, len = s.size();
int left = 0;
memset(cnt, 0, sizeof(cnt));
for(int right = 0 ; right < len; ++right){
int index = s[right] - 0;
++cnt[index];
while(cnt[index] >= 2){
int delIndex = s[left] - 0;
--cnt[delIndex];
++left;
}
ans = max(ans, right - left + 1);
}
return ans;
}
};
Complexity analysis
- Time complexity: O(n). It can be seen that the left and right pointers have moved n times
- Space complexity: O(1)