340. Longest Substring with At Most K Distinct Characters
package LeetCode_340 /** * 340. Longest Substring with At Most K Distinct Characters * (Locked by leetcode) * https://www.lintcode.com/problem/longest-substring-with-at-most-k-distinct-characters/description * * Given a string S, find the length of the longest substring T that contains at most k distinct characters. Example 1: Input: S = "eceba" and k = 3 Output: 4 Explanation: T = "eceb" Example 2: Input: S = "WORLD" and k = 4 Output: 4 Explanation: T = "WORL" or "ORLD" * */ class Solution { /* Sliding Window solution, Time complexity:O(n), Space complexity:O(256) e c e b a ^ i ^ j * occMap: e:1, c:1, b:1 * counter: 3 (when counter > K, need to move i) * best length: 3 * * when moving i: * occMap[s[i]]-- * if (occMap[s[i]]==0){ * counter-- * } * * when moving j: * if (occMap[s[j]])==0{ * counter++ * } * occMap[s[j]]++ * * bestLength = max(bestLength, length of current window)=>max(bestLength, j-i+1) * * counter: how many distinct character we have * * valid window condition: * window contains <= 2 distinct characters = window desirable; * when move i: when the window becomes not desirable (counter>2) * when move j: as long as counter<=2 (as long as window desirable) * * Sliding Window Tech important point: * 1.when to move i, and what have to do when moving i; * 2.when to move j, and what have to do when moving j; * 3.when to update the goal; * */ fun lengthOfLongestSubstringKDistinct(s: String, k: Int): Int { val occMap = IntArray(256) var i = 0 var j = 0 var counter = 0 var bestLength = 0 while (j < s.length) { if (occMap[s[j].toInt()] == 0) { counter++ } occMap[s[j].toInt()]++ //if the count of current distinct character lager than k, //decrease the left pointer and keep tracking the distinct character while (counter > k) { occMap[s[i].toInt()]-- if (occMap[s[i].toInt()] == 0) { counter-- } i++ } bestLength = Math.max(bestLength, j - i + 1) j++ } //print(bestLength) return bestLength } }
标签:
sliding-window
, leetcode
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