LeetCode Top Interview Questions 340. Longest Substring with At Most K Distinct Characters (Java版;

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LeetCode Top Interview Questions 340. Longest Substring with At Most K Distinct Characters (Java版; Hard)

题目描述

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

Example 1:

Input: s = "eceba", k = 2
Output: 3
Explanation: T is "ece" which its length is 3.
Example 2:

Input: s = "aa", k = 1
Output: 2
Explanation: T is "aa" which its length is 2.

第一次做; 跟下面的方法一样, 就是把res换了个位置

class Solution {
    public int lengthOfLongestSubstringKDistinct(String s, int k) {
        if(s==null || s.length()==0 || k==0)
            return 0;
        int left=0, right=0, n=s.length();
        int res = 0;
        HashMap<Character, Integer> map = new HashMap<>();
        //滑动窗口, 双层循环; 外循环中右移right,直至满足条件; 满足条件时, 内循环中右移left,直至不满足条件; 然后重新右移right
        //外循环中右移right
        while(right<n){
            char ch = s.charAt(right);
            map.put(ch, map.getOrDefault(ch,0)+1);
            //内循环中右移left, 缩小滑窗
            while(map.size()>k){
                char ch2 = s.charAt(left);
                map.put(ch2, map.get(ch2)-1);
                if(map.get(ch2)==0)
                    map.remove(ch2);
                left++;
            }
            //换位置之后的res
            res = Math.max(res, right - left + 1);
            right++;
        }
        return res;
    }
}

第一次做; 滑动窗口, 核心:right往右不断扩大窗口到满足(超过)条件, 满足条件时, 内循环中,left不断往右缩小窗口, 直到不满足条件; 不满足条件时继续向右移动right; 细节: res只有在满足条件时才更新, 并且res取历次中的最大值

class Solution {
    public int lengthOfLongestSubstringKDistinct(String s, int k) {
        if(s==null || s.length()==0 || k==0)
            return 0;
        int left=0, right=0, n=s.length();
        int res = 0;
        HashMap<Character, Integer> map = new HashMap<>();
        //滑动窗口, 双层循环; 外循环中右移right,直至满足条件; 满足条件时, 内循环中右移left,直至不满足条件; 然后重新右移right
        //外循环中右移right
        while(right<n){
            char ch = s.charAt(right);
            map.put(ch, map.getOrDefault(ch,0)+1);
            if(map.size()<=k)
                //细节
                res = Math.max(res, right - left + 1);
            //内循环中右移left, 缩小滑窗
            while(map.size()>k){
                char ch2 = s.charAt(left);
                map.put(ch2, map.get(ch2)-1);
                if(map.get(ch2)==0)
                    map.remove(ch2);
                left++;
            }
            right++;
        }
        return res;
    }
}

力扣优秀题解

class Solution:
    def lengthOfLongestSubstringKDistinct(self, s: str, k: int) -> int:
        res = 0
        left = 0
        queue = {}
        for right in range(len(s)):
            queue.setdefault(s[right], 0)
            queue[s[right]] += 1
            while len(queue) > k:
                queue[s[left]] -= 1
                if queue[s[left]] == 0:
                    queue.pop(s[left])
                left += 1
            res = max(res, right - left + 1)
        return res