Longest Substring with at Most k Distinct Characters

原題網址

解題思路：與Minimum Window Substring類似，但此解法只過了四個case，日後找到bug後再來修正。

public int lengthOfLongestSubstringKDistinct(String s, int k) {

    if (s == null || s.length() == 0 || k == 0) {
        return 0;
    }

    int maxLength = 0;
    HashMap<Character, Integer> map = new HashMap<Character, Integer>();

    for (int i = 0, j = 0; i < s.length(); i++) {

        while (j < s.length() && map.size() < k) {

            char cur = s.charAt(j);
            if (map.containsKey(cur)) {
                map.put(cur, map.get(cur) + 1);
            } else {
                map.put(cur, 1);
            }
            j++;
        } 

        if (map.size() <= k) {
            maxLength = Math.max(maxLength, j - i);    
        }

        char charToBeDelete =  s.charAt(i);
        if (map.get(charToBeDelete) > 1) {
            map.put(charToBeDelete, map.get(charToBeDelete) - 1);
        } else {
            map.remove(charToBeDelete);
        }
    }

    return maxLength;
}

九章解法

public int lengthOfLongestSubstringKDistinct(String s, int k) {

    if (s == null || s.length() == 0 || k == 0) {
        return 0;
    }

    int maxLength = 0;
    HashMap<Character, Integer> map = new HashMap<Character, Integer>();

    for (int j = 0, i = 0; j < s.length(); j++) {

        char c = s.charAt(j);
        if (map.containsKey(c)) {
            map.put(c, map.get(c) + 1);
        } else {
            map.put(c, 1);
            while (map.size() > k) {
                char charToBeDelete = s.charAt(i);
                int count = map.get(charToBeDelete);
                if (count > 1) {
                    map.put(charToBeDelete, map.get(charToBeDelete) - 1);
                } else {
                    map.remove(charToBeDelete);
                }
                i++;
            }
        }

        maxLength = Math.max(maxLength, j - i + 1);
    }

    return maxLength;
}