第76題，題目如下:
Given two strings s and t of lengths m and n respectively, return the minimum window
of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "".
The testcases will be generated such that the answer is unique.
Example 1:
Input: s = "ADOBECODEBANC", t = "ABC" Output: "BANC" Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.
Example 2:
Input: s = "a", t = "a" Output: "a" Explanation: The entire string s is the minimum window.
Example 3:
Input: s = "a", t = "aa" Output: "" Explanation: Both 'a's from t must be included in the window. Since the largest window of s only has one 'a', return empty string.
Constraints:
m == s.lengthn == t.length1 <= m, n <= 105sandtconsist of uppercase and lowercase English letters.
Follow up: Could you find an algorithm that runs in O(m + n) time?
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// Q76. Minimum Window Substring
public String minWindow(String s, String t) {
int[] count = new int[128];
int required = t.length();
int bestLeft = –1;
int minLength = s.length() + 1;
for (final char c : t.toCharArray())
++count[c];
for (int l = 0, r = 0; r < s.length(); ++r) {
if (–count[s.charAt(r)] >= 0)
–required;
while (required == 0) {
if (r – l + 1 < minLength) {
bestLeft = l;
minLength = r – l + 1;
}
if (++count[s.charAt(l++)] > 0)
++required;
}
}
return bestLeft == –1 ? “” : s.substring(bestLeft, bestLeft + minLength);
}
