LeetCode - 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：举个例子吧

S="aabcdafac" T="ac"

保存两个指针，分别指向当前已找到的字符串的start和end位置；我们先找到了aabc，这是第一个包含ac的字符串，好的，这个长度为4，start=0，end=3，这时，保持end不动，移动start，发现start=1，end=3可以更短。之后再就不能移start了。

接下来，咱们移动end，发现移到end=4，不是a或c，那就再移，移到end=5，发现是a，这时start还是1，咱们移动start，发现可以一直移到start=3。

接下来再移动end，如此类推

结合代码走一走，就会发现这个思路还是很线性的。

package string;import java.util.HashMap;
import java.util.Map;public class MinimumWindowSubstring {public String minWindow(String s, String t) {int m = s.length();int n = t.length();Map<Character, Integer> found = new HashMap<Character, Integer>();Map<Character, Integer> base = new HashMap<Character, Integer>();for (int i = 0; i < n; ++i) {char c = t.charAt(i);if (base.containsKey(c)) {base.put(c, base.get(c) + 1);} else {base.put(c, 1);found.put(c, 0);}}int count = 0;int ss = -1;int ee = m;for (int end = 0, start = 0; end < m; ++end) {char c = s.charAt(end);if (base.containsKey(c)) {found.put(c, found.get(c) + 1);if (found.get(c) <= base.get(c))++count;if (count == n) {char startChar = s.charAt(start);while (!base.containsKey(startChar) || found.get(startChar) > base.get(startChar)) {if (base.containsKey(startChar))found.put(startChar, found.get(startChar) - 1);startChar = s.charAt(++start);}if (end - start < ee - ss) {ee = end;ss = start;}}}}return ss == -1 ? "" : s.substring(ss, ee + 1);}public static void main(String[] args) {// TODO Auto-generated method stubString S = "ADOBECODEBANC";String T = "ABC";MinimumWindowSubstring m = new MinimumWindowSubstring();System.out.println(m.minWindow(S, T));}}

转载于:https://www.cnblogs.com/null00/p/5094630.html

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