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LeetCode: Minimum Window Substring
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 emtpy string ""
.
If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
地址:https://oj.leetcode.com/problems/minimum-window-substring/
算法:由于申请两个大小为256的数组,用于统计每一个字符串中各字母的个数,然后用isWindow函数来判断是否由hashS表示的字符串包含由hashT表示的字符串。首先,从S开始找到第一在T中出现的字母位置,记为i,初始化start为i,其中count表示在字符串S[start~i]之间出现在字符串T的字符个数。在每次进入循环时都判断S[start~i]是否可以包含T,如果包含,则根据情况更新最小窗口,并且更新start为下一个出现在字符串T中位置(因为最小的窗口的其实位置一定会出现在字符串T中),注意还要将改变hashS让其表示当前的S[start~i](注意i还是不变);如果不包含,则改变i为下一个出现在字符串T中的位置,并且同时更新对应的hashS。代码:
1 class Solution {
2 public:
3 string minWindow(string S, string T) {
4 int slen = S.size();
5 int tlen = T.size();
6 if(slen <= 0 || tlen <= 0) return string();
7 vector<int> hashT(256,0);
8 vector<int> hashS(256,0);
9 for(int i = 0; i < tlen; ++i)
10 ++hashT[T[i]];
11 int minNum = slen + 1;
12 int minStart = 0, minEnd = 0;
13 int i = 0;
14 while(i < slen && !hashT[S[i]]){
15 ++i;
16 }
17 if(i >= slen) return string();
18 int start = i;
19 int count = 1;
20 ++hashS[S[i]];
21 while(i < slen){
22 if(isWindow(hashT,hashS,count,tlen)){
23 if(minNum > i-start+1){
24 minNum = i-start+1;
25 minStart = start;
26 minEnd = i;
27 }
28 --count;
29 --hashS[S[start]];
30 ++start;
31 while(start<slen && !hashT[S[start]]) ++start;
32
33 }else{
34 ++i;
35 while(i<slen && !hashT[S[i]]) ++i;
36 if(i < slen){
37 ++count;
38 ++hashS[S[i]];
39 }
40 }
41 }
42 if(minNum == slen + 1){
43 return string();
44 }
45 return S.substr(minStart,minNum);
46 }
47 bool isWindow(vector<int> &hashT, vector<int> &hashS, int count ,int tlen){
48 if(count < tlen)
49 return false;
50 for(int i = 0; i < 256; ++i){
51 if(hashS[i] < hashT[i]){
52 return false;
53 }
54 }
55 return true;
56 }
57 };
LeetCode: Minimum Window Substring
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原文地址:http://www.cnblogs.com/boostable/p/leetcode_minimum_window_substring.html