【LeetCode】Minimum Window Substring

题目描述：

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.

思路是双指针分别记录窗口的左右端点，右指针不断右移，每当遇到T中的元素时收缩左端点到最小。

那么这题的主要问题就是如何收缩左端点。一开始想的有点麻烦，用map<char, vector>记录下每个元素的坐标，当右端点遇到T中元素时将最小的去掉，再push进当前坐标。后来想到，因为不考虑顺序，只要记录窗口内元素的个数是否比T中对应元素的个数多就可以了，又重新写了下。

两种方法都可以AC，时间也差不多，可能还是测试数据不够大吧，因为vector中删除操作为O（n）复杂度，数组占用空间也更多，数据再大一点第一种方法可能就通不过了。

代码如下：

class Solution {public:string minWindowII(string S, string T) {unordered_map<char, int> count, curr;int left(0), right(S.length() - 1), l(-1);for (int i = 0; i < T.length(); i++)count[T[i]]++;curr = count;for (int i = 0; i < S.length(); i++){if (count.count(S[i])){if (l < 0)l = i;count[S[i]]--;curr[S[i]]--;if (count[S[i]] == 0)count.erase(S[i]);}else if (curr.count(S[i])){curr[S[i]]--;while (!curr.count(S[l]) || curr[S[l]] < 0){if (curr.count(S[l]))curr[S[l]]++;l++;}}if (count.empty() && i - l < right - left){left = l;right = i;}}if (!count.empty())return "";return S.substr(left, right - left + 1);}

string minWindowI(string S, string T) {unordered_map<char, int> count;unordered_map<char, vector<int>> index;int left(0), right(S.length() - 1);for (int i = 0; i < T.length(); i++)count[T[i]]++;int l(-1);for (int i = 0; i < S.length(); i++){if (count.count(S[i])){if (l < 0)l = i;index[S[i]].push_back(i);count[S[i]]--;if (!count[S[i]])count.erase(S[i]);}else if (index.count(S[i])){vector<int> *v = &index[S[i]];v->erase(v->begin());v->push_back(i);if (S[i] == S[l])do{l++;} while (!(index.count(S[l]) && index[S[l]][0] == l));}if (count.empty() && i - l < right - left){left = l;right = i;}}if (!count.empty())return "";return S.substr(left, right - left + 1);}};