[leetcode] Minimum Window Substring

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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.

https://oj.leetcode.com/problems/minimum-window-substring/

 

思路：視窗移動法。 詳見這裡

public class Solution {    public String minWindow(String S, String T) {        if (S == null || T == null)            return "";        int[] needToFind = new int[256];        int[] found = new int[256];        for (int i = 0; i < T.length(); i++)            needToFind[T.charAt(i)]++;        String result = "";        int begin = 0, end = 0;        int minWin = Integer.MAX_VALUE;        int count = 0;        for (end = 0; end < S.length(); end++) {            if (needToFind[S.charAt(end)] == 0)                continue;            char ch = S.charAt(end);            found[ch]++;            if (found[ch] <= needToFind[ch])                count++;                        if (count == T.length()) {                //move the begin pointer while the constrain meets                while (begin < S.length()                        && (found[S.charAt(begin)] > needToFind[S.charAt(begin)] || needToFind[S.charAt(begin)] == 0)) {                    if (found[S.charAt(begin)] > 0)                        found[S.charAt(begin)]--;                    begin++;                }                //update the min according to the current window                if (end - begin + 1 < minWin) {                    minWin = end - begin + 1;                    result = S.substring(begin, end + 1);                }            }        }        return result;    }    public static void main(String[] args) {        System.out.println(new Solution().minWindow("ADOBECODEBANC", "ABC"));    }}

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