[LeetCode][Java] Minimum Window Substring

题目：

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.

题意：

给定一个字符串S 和一个字符串T,在字符串S中找出最小的窗体，这个窗体能够包括T中的全部的字符，时间复杂度要求为O(n).

举个样例：

S =

"ADOBECODEBANC"

T =

"ABC"

最下的窗体为"BANC"

.

算法分析：

// 双指针思想，尾指针不断往后扫，当扫到有一个窗体包括了全部T的字符，然后再收缩头指针。直到不能再收缩为止。

// 最后记录全部可能的情况中窗体最小的

AC代码：

public class Solution
{
public String minWindow(String S, String T)
{
HashMap<Character, Integer> hasFound = new HashMap<Character, Integer>();
HashMap<Character, Integer> needToFind = new HashMap<Character, Integer>();

for (int i = 0; i < T.length(); i++)
{
hasFound.put(T.charAt(i), 0);
if (needToFind.containsKey(T.charAt(i)))
{
needToFind.put(T.charAt(i), needToFind.get(T.charAt(i)) + 1);
}
else
{
needToFind.put(T.charAt(i), 1);
}
}
int begin = 0;
int minWindowSize = S.length();
String retString = "";

int count = 0;

for (int end = 0; end < S.length(); end++)
{
Character end_c = S.charAt(end);
if (needToFind.containsKey(end_c))
{
hasFound.put(end_c, hasFound.get(end_c) + 1);
if (hasFound.get(end_c) <= needToFind.get(end_c))
{
count++;
}
if (count == T.length())
{
while ((!needToFind.containsKey(S.charAt(begin)))||(hasFound.get(S.charAt(begin)) > needToFind.get(S.charAt(begin))))
{
if (needToFind.containsKey(S.charAt(begin)))
{
hasFound.put(S.charAt(begin),hasFound.get(S.charAt(begin)) - 1);
}
begin++;
}
if ((end - begin + 1) <= minWindowSize)
{
minWindowSize = end - begin + 1;
retString = S.substring(begin, end + 1);
}
}
}
}
return retString;
}
}