LeetCode 76. Minimum Window Substring

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本文介绍了一种在字符串S中寻找包含字符串T所有字符的最短子串的算法。通过使用滑动窗口技术和哈希映射，该算法能在O(n)的时间复杂度内找到符合条件的最小子串。

76. Minimum Window Substring

Description

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.

    class Solution {
        public String minWindow(String s, String t) {
            if (s.isEmpty() || t.isEmpty() || s.length() < t.length())
                return "";
            int minlen = s.length();
            int minstart = 0;
            int minend = s.length()-1;
            HashMap<Character, Integer> require = new HashMap<Character, Integer>();

            for (char c : t.toCharArray()) {
                require.put(c, require.containsKey(c) ? require.get(c) + 1 : 1);
            }

            int count = t.length();
            int li = 0;

            for (int i = 0; i < s.length(); i++) {
                char c = s.charAt(i);

                if (require.containsKey(c)) {
                    if (require.get(c) > 0)
                        count--;
                    require.put(c, require.get(c) - 1);
                }

                if (count == 0) {
                    char cli = s.charAt(li);
                    while (!require.containsKey(cli) || require.get(cli) < 0) {
                        if (require.containsKey(cli)) {
                            require.put(cli, require.get(cli) + 1);
                        }
                        li++;
                        cli = s.charAt(li);
                    }

                    if (minlen > i - li + 1) {
                        minstart = li;
                        minend = i;
                        minlen = i - li + 1;
                    }
                }
            }
            if (count != 0) {
                return "";
            }
            return s.substring(minstart, minend + 1);
        }