Leetcode: Longest Substring with At Most K Distinct Characters && Summary: Window做法两种思路总结

Given a string, find the length of the longest substring T that contains at most k distinct characters.

For example, Given s = “eceba” and k = 2,

T is "ece" which its length is 3.

我的做法：维护一个window，r移动到超出k distinct character限制是更新max，然后移动l使distinct character <=k; 这种做法更新max只发生在超出限制的时候，有可能永远都没有超出限制，所以while loop完了之后要补上一个max的更新

public class Solution {
    public int lengthOfLongestSubstringKDistinct(String s, int k) {
        int l=0, r=0; 
        HashMap<Character, Integer> map = new HashMap<Character, Integer>();
        int longest = 0;
        if (s==null || s.length()==0 || k<=0) return longest;
        while (r < s.length()) {
            char cur = s.charAt(r);
            map.put(cur, map.getOrDefault(cur, 0) + 1);
            if (map.size() > k) {
                longest = Math.max(longest, r-l);
                while (map.size() > k) {
                    char rmv = s.charAt(l);
                    if (map.get(rmv) == 1) map.remove(rmv);
                    else map.put(rmv, map.get(rmv)-1);
                    l++;
                }
            }
            r++;
        }
        longest = Math.max(longest, r-l);
        return longest;
    }
}

 

别人非常棒的做法：sliding window，while loop里面每次循环都会尝试更新max，所以每次更新之前都是把l, r调整到合适的位置，也就是说，一旦r移动到超出K distinct char限制，l也要更新使k distinct char重新满足，l更新是在max更新之前

public class Solution {
    public int lengthOfLongestSubstringKDistinct(String s, int k) {
        int[] count = new int[256];
        int num = 0, i = 0, res = 0; //num is # of distinct char, i is left edge, j is right edge
        for (int j = 0; j < s.length(); j++) {
            if (count[s.charAt(j)]++ == 0) num++;
            if (num > k) {
                while (--count[s.charAt(i++)] > 0);
                num--;
            }
            res = Math.max(res, j - i + 1);
        }
        return res;
    }
}