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.

双指针的思想，先找到一个包含T的窗口，然后在满足条件（包含T）的情况下向后移动头指针，如果不能再移动则继续移动尾指针。这里用一个计数器统计字符串S中字符串T内字符出现的次数 - 可以是重复字符，只要总次数小于T中的重复次数 - 如果等于T的长度表明已经包含T。

class Solution {
public:
string minWindow(string S, string T) {
int chars_to_find[256] = {0};
for (int i = 0; i < T.size(); ++i) {
++chars_to_find[T[i]];
}

int ret, minlen = S.size() + 1;
int count = 0;
int has_found[256] = {0};
for (int i = 0, start = 0; i < S.size(); ++i) {
if (chars_to_find[S[i]] == 0) {
continue;
}

++has_found[S[i]];
if (has_found[S[i]] <= chars_to_find[S[i]]) {
++count;
}

if (count == T.size()) {
// have found all the characters in T
while (chars_to_find[S[start]] == 0 ||
has_found[S[start]] > chars_to_find[S[start]]) {
if (has_found[S[start]] > chars_to_find[S[start]]) {
--has_found[S[start]];
}
++start;
}

if (i - start + 1 < minlen) {
ret = start;
minlen = i - start + 1;
}
}
}

return minlen == S.size() + 1 ? "" : S.substr(ret, minlen);
}
};

====================第二次===================

第二次也写了一个半小时啊。。。

class Solution {
public:
string minWindow(string S, string T) {
int expected_chars[256] = {0};
int total_count = T.size();
for (int i = 0; i < total_count; ++i) {
++expected_chars[T[i]];
}

int found_count = 0;
int found_chars[256] = {0};
int min_length = S.size() + 1;
int min_start = -1;
for (int i = 0, start = 0; i < S.size(); ++i) {
if (expected_chars[S[i]] == 0) {
continue;
}

++found_chars[S[i]];
if (found_chars[S[i]] <= expected_chars[S[i]]) {
++found_count;
while (found_count == total_count) {
while (expected_chars[S[start]] == 0) {
++start;
}
int length = i - start + 1;
if (length < min_length) {
min_length = length;
min_start = start;
}

--found_chars[S[start]];
if (found_chars[S[start]] < expected_chars[S[start]]) {
--found_count;
}
++start;
}
}
}

return min_start == -1 ? "" : S.substr(min_start, min_length);
}
};