【Leetcode】Minimum Window Substring

题目链接：https://leetcode.com/problems/minimum-window-substring/

题目：

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 empty string 

""

.

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

思路：

时间太久 我也忘了咋做的了= =有时间再补

算法：

public String minWindow(String s, String t) {
String result = "";
StringBuffer res = new StringBuffer();
HashMap<String, Integer> tm = new HashMap<String, Integer>();// 维护窗口内需要的t字符的个数，如果某字符值小于0表示该窗口有多于需要的字符个数
int minLen = Integer.MAX_VALUE, start = 0, end = 0, size = t.length();// size
// 表示剩余要匹配t的字符个数

for (int i = 0; i < t.length(); i++) { // 初始化
String key = t.charAt(i) + "";
if (tm.containsKey(key)) {
tm.put(key, tm.get(key) + 1);
} else {
tm.put(key, 1);
}
}

while (end < s.length()) {
while (end < s.length() && size > 0) {
String key = s.charAt(end) + "";
if (tm.containsKey(key)) {
if (tm.get(key) > 0) { // 如果需要该字符
size--;
}
tm.put(key, tm.get(key) - 1);
}
res.append(key);
end++;
}// 已经找到一个window包含了t所有字符
while (size == 0 && start < end) {
if (minLen > end - start) { // 更新窗口大小
minLen = end - start;
result = s.substring(start, end);
}
String key = s.charAt(start) + "";
if (tm.containsKey(key)) {
if (tm.get(key) == 0) { // 如果窗口删除该字符，则跟t的匹配缺少该字符
size++;
}
tm.put(key, tm.get(key) + 1);
res.deleteCharAt(0);
}
start++;
}
}
return result;
}