0003. Longest Substring Without Repeating Characters (M)

Longest Substring Without Repeating Characters (M)

题目

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.

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题意

给定一个字符串，找到一个子串，使得子串内没有重复的字符，求该子串的最大长度。

思路

普通解法直接暴力三重循环(O(N^3))。

较优解法是使用一个“可伸缩”的数组框在字符串上滑动，先向右移动右端点，判断右端点的字符在不在被“伸缩数组”框住的子串中，如果重复则不断向右移动左端点，直到将重复的字符排除在“伸缩数组”外。这样相当于字符串中的每一个字符都要走一遍，复杂度(O(2N) = O(N))。

对上述方法的一个优化是，用一个数组记录字符串中每一个字符对应的索引位置，这样在右端点字符重复时，左端点可以直接跳到子串中字符重复的地方，不需要一步一步向右移动左端点。这样只需要右端点走一遍字符串即可，复杂度直接为(O(N))。

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代码实现

Java

Set

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int left = 0;
        // right指向子串的右边一个字符
        int right = 0;
        int maxLen = 0;
        Set<Character> set = new HashSet<>();
        while (right < s.length()) {
            if (set.contains(s.charAt(right))) {
                set.remove(s.charAt(left++));
            } else {
                set.add(s.charAt(right++));
                maxLen = Math.max(right - left, maxLen);
            }
        }
        return maxLen;
    }
}

Hash优化

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int left = 0;
        // right指向子串的最后一个字符
        int right = 0;
        int maxLen = 0;
        // 下标记录字符，值记录该字符在字符串中对应的索引+1
        // 实际上记录的是左端点应该跳转到的位置
        int[] hash = new int[128];
        while (right < s.length()) {
            left = Math.max(left, hash[s.charAt(right)]);
            maxLen = Math.max(maxLen, right - left + 1);
            hash[s.charAt(right)] = ++right;
        }
        return maxLen;
    }
}

JavaScript

Set

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLongestSubstring = function (s) {
  let left = 0
  let right = 0
  let max = 0
  let set = new Set()

  while (right < s.length) {
    if (set.has(s[right])) {
      set.delete(s[left++])
    } else {
      set.add(s[right++])
      max = Math.max(right - left, max)
    }
  }

  return max
}

Hash优化

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLongestSubstring = function (s) {
  let left = 0
  let right = 0
  let max = 0
  let hash = new Map()

  while (right < s.length) {
    let index = hash.get(s[right]) !== undefined ? hash.get(s[right]) : -1
    left = Math.max(left, index + 1)
    max = Math.max(max, right - left + 1)
    hash.set(s[right], right++)
  }

  return max
}