159. Longest Substring with At Most Two Distinct Characters

最后更新

二刷

08-Jan-17

回头看了下一刷的，用的map,应该是int[256]的意思，后面没仔细看cuz whatever I was doing at that time.. wasnt good

做法和LC 76非常像，用2 Pointers + 计数来判断是否满足。

这里“有效读取”的判断标准变成了 count[s.charAt(someIndex)]是否从0递增，和每个循环最后它是否递减回0，以此判断dinstinct是否有变化，其实这个比76的有效读取要稍微好理解一些。

Time: O(N)

Space: Constant space..

public class Solution {
    public int lengthOfLongestSubstringTwoDistinct(String s) {
        if (s.length() <= 2) return s.length();

        int[] count = new int[256];
        int temp = 0;
        int right = 0;
        int maxLength = -1;
        for (int left = 0; left < s.length(); left ++) {
            while (right < s.length()) {
                if (temp == 2 && count[s.charAt(right)] == 0) break;
                if (++count[s.charAt(right++)] == 1) {
                    temp ++;
                }
            }

            if (right - left > maxLength) {
                maxLength = right - left;
            }

            if (--count[s.charAt(left)] == 0) {
                temp --;
            }
        }
        return maxLength;
    }
}

---

一刷

18-Dec-2016

怎么和上一题一样的。。。

map快。。

import java.util.Hashtable;
public class Solution
{
    public int lengthOfLongestSubstringTwoDistinct(String s)
    {
        if(s.length() <= 2) return s.length();

        Hashtable<Character,Integer> table = new Hashtable<Character,Integer>();

        int temp = 0;

        int l = 0;

        int res = 1;

        for(int i = 0; i < s.length();i++)
        {
            char c = s.charAt(i);
            if(!table.containsKey(c)) temp++;
            table.put(c,i);
            if(temp > 2)
            {
                temp--;
                Iterator iter = table.keySet().iterator();
                char iterC = c;
                int index = i;

                while(iter.hasNext())
                {
                    char tempC = (char)iter.next();
                    if(table.get(tempC) < index)
                    {
                        index = table.get(tempC);
                        iterC = tempC;
                    }
                }

                table.remove(iterC);
                l = index + 1;
            }

            res = Math.max(res,i+1-l);
        }
        return res;
    }
}