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.
*/
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
/**
 * @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++;
    }
};

  

posted @   johnny_zhao  阅读(87)  评论(0编辑  收藏  举报
编辑推荐:
· AI与.NET技术实操系列(二):开始使用ML.NET
· 记一次.NET内存居高不下排查解决与启示
· 探究高空视频全景AR技术的实现原理
· 理解Rust引用及其生命周期标识(上)
· 浏览器原生「磁吸」效果!Anchor Positioning 锚点定位神器解析
阅读排行:
· DeepSeek 开源周回顾「GitHub 热点速览」
· 物流快递公司核心技术能力-地址解析分单基础技术分享
· .NET 10首个预览版发布:重大改进与新特性概览!
· AI与.NET技术实操系列(二):开始使用ML.NET
· .NET10 - 预览版1新功能体验(一)
点击右上角即可分享
微信分享提示