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 empty string “”.
If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
分析
在字符串 S 中查找最小的窗口，使窗口中全部包含字符串T 中的字符，（不按顺序）。
需要注意的地方：
- T 中的字符可以重复，窗口需要重复包含。
- 测试 实验中的是SCII字符 ，所以可以用128位数组代替 hash table。
AC代码
class Solution {
public:
string minWindow(string S, string T) {
if (S.empty() || T.empty())
{
return "";
}
int count = T.size();
int require[128] = { 0 };
bool chSet[128] = { false };
for (int i = 0; i < count; ++i)
{
require[T[i]]++;
chSet[T[i]] = true;
}
int i = -1;
int j = 0;
int minLen = INT_MAX;
int minIdx = 0;
while (i < (int)S.size() && j < (int)S.size())
{
if (count)
{
i++;
require[S[i]]--;
if (chSet[S[i]] && require[S[i]] >= 0)
{
count--;
}
}//if
else
{
if (minLen > i - j + 1)
{
minLen = i - j + 1;
minIdx = j;
}
require[S[j]]++;
if (chSet[S[j]] && require[S[j]] > 0)
{
count++;
}
j++;
}//else
}//while
if (minLen == INT_MAX)
{
return "";
}
return S.substr(minIdx, minLen);
}
};