[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.

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Hash Table Two Pointers String

思路：

双指针，动态维护一个区间。尾指针不断往后扫，当扫到有一个窗口包含了所有 T 的字符后，
然后再收缩头指针，直到不能再收缩为止。最后记录所有可能的情况中窗口最小的

时间复杂度 O(n)，空间复杂度 O(1)

class Solution {
public:
string minWindow(string S, string T)
{
if(S.empty() || T.empty() || S.size() < T.size())
return string();

vector<int> expect(256, 0);
vector<int> appear(256, 0);

for(int i = 0; i < T.size(); i++)
{
expect[T[i]] ++;
}

int minWidth = INT_MAX, min_start = 0; //
int win_start = 0;
int appearCharCnt = 0;
for(int win_end = 0; win_end < S.size(); win_end++)
{
if(expect[S[win_end]] > 0) // this char is part of T
{
appear[S[win_end]]++;
if(appear[S[win_end]] <= expect[S[win_end]])
appearCharCnt ++;
}
//cout << "appearCharCnt\t" <<appearCharCnt<< endl;
if(appearCharCnt == T.size())
{
// shrink the start
while (appear[S[win_start]] > expect[S[win_start]]
|| expect[S[win_start]] == 0) {
appear[S[win_start]]--;
win_start++;
}
if ((win_end - win_start + 1) < minWidth) {
minWidth = win_end - win_start + 1;
min_start = win_start;
//cout << "min_start\t" <<min_start << endl;
//cout << "min_width\t" <<minWidth<< endl;
}
}
}

if (minWidth == INT_MAX)
return "";
else
return S.substr(min_start, minWidth);

}
};

精简一下条件判断

class Solution {
public:
string minWindow(string S, string T)
{
if(S.empty() || T.empty() || S.size() < T.size())
return string();

vector<int> expect(256, 0);
vector<int> appear(256, 0);

for(int i = 0; i < T.size(); i++)
{
expect[T[i]] ++;
}

int minWidth = INT_MAX, min_start = 0;
int win_start = 0;
int appearCharCnt = 0;
for(int win_end = 0; win_end < S.size(); win_end++)
{
appear[S[win_end]]++;
if(appear[S[win_end]] <= expect[S[win_end]])
appearCharCnt ++;
//cout << "appearCharCnt\t" <<appearCharCnt<< endl;
if(appearCharCnt == T.size())
{
// shrink the win start
while (appear[S[win_start]] > expect[S[win_start]]
) {
appear[S[win_start]]--;
win_start++;
}
if ((win_end - win_start + 1) < minWidth) {
minWidth = win_end - win_start + 1;
min_start = win_start;
//cout << "min_start\t" <<min_start << endl;
//cout << "min_width\t" <<minWidth<< endl;
}
}
}

if (minWidth == INT_MAX)
return "";
else
return S.substr(min_start, minWidth);

}
};