zoukankan      html  css  js  c++  java
  • 76. Minimum Window Substring

    /**
    * 76. Minimum Window Substring
    * https://leetcode.com/problems/minimum-window-substring/description/
    *
    * Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
    Example:
    Input: S = "ADOBECODEBANC", T = "ABC"
    Output: "BANC"
    Note:
    If there is no such window in S that covers all characters in T, return the empty string "".
    If there is such window, you are guaranteed that there will always be only one unique minimum window in S.
    */
    /**
     * @param {string} s
     * @param {string} t
     * @return {string}
     */
    var minWindow = function (s, t) {
        if (s == "" || t == "") return "";
        let matchCount = 0, result = "";
        let arrT = [], arrS = [];
        //set up arrT and arrS
        for (let c of t) {
            if (arrT[c] == NaN || arrT[c] == undefined)
                arrT[c] = 1;
            else
                arrT[c]++;
        }
        for (let c of s) {
            if (arrS[c] == undefined)
                arrS[c] = 0;
        }
        //console.log(arrS);
        let left = findNextStr(0, s, arrT);
        if (left == s.length)
            return "";
        let right = left;
        //start scan. 
        //use right pointer to check each letter
        while (right < s.length) {
            let rightChar = s.charAt(right);
            if (arrS[rightChar] < arrT[rightChar])
                matchCount++;
            arrS[rightChar]++;
            while (left < s.length && matchCount == t.length) {
                if (result == "" || result.length > right - left + 1) {
                    //如果result为空或者找到比当前result更短的
                    result = s.substring(left, right + 1);
                }
                //start to move left pointer
                let leftChar = s.charAt(left);
                if (arrS[leftChar] <= arrT[leftChar])
                    matchCount--;
                arrS[leftChar]--;
                //left pointer move to next vaild character
                left = findNextStr(left + 1, s, arrT);
            }
            //right pointer also move to next vaild character
            //right = findNextStr(right + 1, s, arrT);
            //i changed to right++, fast than right = findNextStr(right + 1, s, arrT);
            right++;
        }
        //console.log(result);
        return result
    };
    
    var findNextStr = function (start, s, arrT) {
        while (start < s.length) {
            let c = s.charAt(start);
            if (arrT[c] > 0)
                return start;
            start++;
        }
    };
    

      

  • 相关阅读:
    !JS实战之随机像素图
    bgp选路原则【第二部】
    BGP基础【第三部】
    【★】KMP算法完整教程
    ★如何引导客户需求?几个经…
    html标签缺省(自带)样式大全
    Web颜色对照表大全
    PS各个工具的字母快捷键和英…
    色相、明度及饱和度
    用webgl打造自己的3D迷宫游戏
  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/10217748.html
Copyright © 2011-2022 走看看