LeetCode 340. Longest Substring with At Most K Distinct Characters（最长字串）

最新推荐文章于 2022-02-18 22:52:34 发布

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本文介绍了一种解决LeetCode上最长包含最多K种不同字符子串问题的方法。通过使用直方图统计字符出现次数，并维护一个滑动窗口来确保窗口内字符种类不超过K种，从而找到符合条件的最长子串长度。

原题网址：https://leetcode.com/problems/longest-substring-with-at-most-k-distinct-characters/

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.

方法：使用直方图统计字符出现次数，如果是首次出现则distinct字符数加1.

public class Solution {
    public int lengthOfLongestSubstringKDistinct(String s, int k) {
        if (s == null) return 0;
        int from = 0;
        int[] histogram = new int[256];
        int distinct = 0;
        char[] sa = s.toCharArray();
        int max = 0;
        for(int i=0; i<sa.length; i++) {
            if (histogram[sa[i]] == 0) distinct ++;
            histogram[sa[i]] ++;
            while (distinct > k) {
                histogram[sa[from]] --;
                if (histogram[sa[from]] == 0) distinct --;
                from ++;
            }
            max = Math.max(max, i-from+1);
        }
        return max;
    }
}

另一种实现：

public class Solution {
    public int lengthOfLongestSubstringKDistinct(String s, int k) {
        int from = 0;
        int max = 0;
        char[] sa = s.toCharArray();
        int[] f = new int[256];
        int d = 0;
        for(int i=0; i<sa.length; i++) {
            if (f[sa[i]]++ == 0) d++;
            while (d>k && from<=i) {
                if (f[sa[from]]==1) d--;
                f[sa[from++]]--;
            }
            max = Math.max(max, i-from+1);
        }
        return max;
    }
}

进一步简化：

public class Solution {
    public int lengthOfLongestSubstringKDistinct(String s, int k) {
        if (k <= 0) return 0;
        int[] f = new int[256];
        int distinct = 0;
        char[] sa = s.toCharArray();
        int from = 0;
        int max = 0;
        for(int i=0; i<sa.length; i++) {
            if (f[sa[i]] ++ == 0) distinct ++;
            while (distinct > k) {
                if (--f[sa[from++]] == 0) distinct --;
            }
            max = Math.max(max, i-from+1);
        }
        return max;
    }
}