LeetCode76 Minimum Window Substring（两种解法）

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.

 

这道题的要求是要在O(n)的时间度里实现找到这个最小窗口字串，那么暴力搜索Brute Force肯定是不能用的，我们可以考虑哈希表，其中key是T中的字符，value是该字符出现的次数。

- 我们最开始先扫描一遍T，把对应的字符及其出现的次数存到哈希表中。

- 然后开始遍历S，遇到T中的字符，就把对应的哈希表中的value减一，直到包含了T中的所有的字符，纪录一个字串并更新最小字串值。

- 将子窗口的左边界向右移，略掉不在T中的字符，如果某个在T中的字符出现的次数大于哈希表中的value，则也可以跳过该字符。

解法一：

class Solution {
public:
string minWindow(string S, string T) {
if (T.size() > S.size()) return "";
string res = "";
int left = 0, count = 0, minLen = S.size() + 1;
unordered_map<char, int> m;
for (int i = 0; i < T.size(); ++i) {
if (m.find(T[i]) != m.end()) ++m[T[i]];
else m[T[i]] = 1;
}
for (int right = 0; right < S.size(); ++right) {
if (m.find(S[right]) != m.end()) {
--m[S[right]];
if (m[S[right]] >= 0) ++count;
while (count == T.size()) {
if (right - left + 1 < minLen) {
minLen = right - left + 1;
res = S.substr(left, minLen);
}
if (m.find(S[left]) != m.end()) {
++m[S[left]];
if (m[S[left]] > 0) --count;
}
++left;
}
}
}
return res;
}
};LeetCode提交时间为26ms。

解法2：

LeetCode提交上最快的cpp解法，6ms，代码如下：
class Solution {
public:

string minWindow(string s, string t) {
vector<int> map(256, 0);
for (int i = 0; i < t.size(); i++) {
map[t[i]]++;
}

int start = 0;
int end = 0;
int count = t.size();
int d = INT_MAX;
int head = 0;
while(end < s.size()) {
// reduce one anyway, if original map[i] is
// greater than zero, that means a char match
// count--
if (map[s[end++]]-- > 0) {
count--;
}

// full match
while (count == 0) {
if (end - start < d) {
// head is only updated when
// end - start is less
d = end - (head = start);
}

// it will be greater than 0, so it is the t character
if (map[s[start++]]++ == 0) {
count++;
}
}
}

if (d == INT_MAX) {
return "";
} else {
return s.substr(head, d);
}
}
};