java_LeetCode_3--Longest Substring Without Repeating Characters

题目;

Given a string, find the length of the longest substring without repeating characters.

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", 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.

首先自己的解法，强行整了动态规划，dp[i]表示s[0...i]的最长无重复子串，时间复杂度O(n*n)，很高

public int lengthOfLongestSubstring(String s) {
		if(s==null || s.length()==0)  //注意特殊情况处理
			return 0;
		int curLen = 0;
		int max = 0;
		int len = s.length();
		int dp[] = new int[len];
		dp[0] = 1;
		Set<Character> set = new HashSet<Character>();
		for (int i = 1; i < s.length(); i++) {
			int j = 0;
			for(j=i;j>=0;j--) {
				if(!set.contains(s.charAt(j))) {
					set.add(s.charAt(j));
				} else {
					dp[i] = Math.max(dp[i-1], set.size());
					set.clear();
					break;
				}
			}
			if(j<0) {
				dp[i] = i+1;
			}
		}
		for(int t : dp) {
			max = max<t ? t : max;
		}
		return max;
	}

O(2n)解法：s[i....j]的滑动窗口

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length();
        Set<Character> set = new HashSet<>();
        int ans = 0, i = 0, j = 0;
        while (i < n && j < n) {
            // try to extend the range [i, j]
            if (!set.contains(s.charAt(j))){
                set.add(s.charAt(j++));
                ans = Math.max(ans, j - i);
            }
            else {
                set.remove(s.charAt(i++));
            }
        }
        return ans;
    }
}

O(n)解法

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length(), ans = 0;
        Map<Character, Integer> map = new HashMap<>(); // current index of character
        // try to extend the range [i, j]
        for (int j = 0, i = 0; j < n; j++) {
            if (map.containsKey(s.charAt(j))) {
                i = Math.max(map.get(s.charAt(j)), i);
            }
            ans = Math.max(ans, j - i + 1);
            map.put(s.charAt(j), j + 1);
        }
        return ans;
    }
}

Assuming ASCII 128：数组替换hash表

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length(), ans = 0;
        int[] index = new int[128]; // current index of character
        // try to extend the range [i, j]
        for (int j = 0, i = 0; j < n; j++) {
            i = Math.max(index[s.charAt(j)], i);
            ans = Math.max(ans, j - i + 1);
            index[s.charAt(j)] = j + 1;
        }
        return ans;
    }
}