LeetCode -- Minimum Window Substring

题目描述：
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).


For example,
S = "ADOBECODEBANC"
T = "ABC"
Minimum window is "BANC".
Note:
If there is no such window in S that covers all characters in T, return the empty string "".


If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.


本题就是要在串S中找到一个字串S[i...k]，其中i,k∈[0,n) ，n为S的长度，使得S[i...k]包含字符串T中的所有字符。




思路：
1. 使用hashT来保存每个字符出现的次数
2. 初始化hashS（将t的每个字符添加到hashS并将值初始化为0，即出现次数都为0），遍历一次s（i∈[0,n)），对于s[i]：
2.1 如果s[i]在hashT中出现，将hashS[s[i]]累加，判断hashS[s[i]]与hashT[s[i]]是否相等，即判断此时是否完成了一次s[i]的映射，如果相等，将mapped变量（初始化为0）++
2.2 当完成了t中所有字符的映射时，即mapped = t中字符个数（不包含重复）时：
从当前窗口最左边left(初始化为0)开始判断：
如果hashT不包含s[left]：直接left++
如果hashT包含s[left]并且hashS[s[left]] 大于hashT[s[left]]，说明对s[left]的映射数量超出了所需要的个数hashT[s[left]]，此时也可以将left++。
此时判断left到i的距离是否小于当前最小窗口，小于则更新窗口即可。




本题的实现参考了以下链接：
https://github.com/yuzhangcmu/LeetCode/blob/master/string/MinWindow.java
http://www.programcreek.com/2014/05/leetcode-minimum-window-substring-java/




实现代码：

public class Solution {    public string MinWindow(string s, string t)     {        // 1. save t[i] into hash and get count of unique char in t    var countT = 0;    var hashT = new Dictionary<char, int>();    for(var i = 0;i < t.Length; i++){    if(!hashT.ContainsKey(t[i])){    hashT.Add(t[i], 1);    countT++;    }    else{    hashT[t[i]]++;    }    }        // 2. init hashS     var hashS = new Dictionary<char, int>();    for(var i = 0;i < s.Length; i++){    if(!hashS.ContainsKey(s[i])){    hashS.Add(s[i], 0);    }    }        var mapped = 0; // without duplicate (say 'bccd' , mapped here means 'bcd')    var left = 0;    var minLen = s.Length;    var result = s;        var found = false;    for (int i = 0; i < s.Length; i++) {    char c = s[i];    if (hashT.ContainsKey(c)) {    hashS[c]++;    // we have done mapping for s[i], increase mapped count for it    if (hashT[c] == hashS[c]) {    mapped++;     }    }        if (mapped == countT)     {    found = true;    var leftC = s[left];    // if first char(say c) is not include in t , or count of c in hashS is more than in hashT then: hashS[c] --     // set left++                while (!hashT.ContainsKey(leftC) || hashS[leftC] > hashT[leftC]){                    if (hashT.ContainsKey(leftC) && hashS[leftC] > hashT[leftC]){    hashS[leftC]--;    }                                            left++;                    leftC = s[left];                }                     if (i - left + 1< minLen) {                    result = s.Substring(left, i - left + 1);                    minLen = i - left + 1;                }    }        }        return !found ? "" : result;    }}