[LintCode/LeetCode] Minimum Window Substring

Corwien 发布于2019-08-14 15:43 / 2489人阅读

Problem

Given a string source and a string target, find the minimum window in source which will contain all the characters in target.

Notice

If there is no such window in source that covers all characters in target, return the emtpy string "".

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

Clarification

Should the characters in minimum window has the same order in target?

Not necessary.

Example

For source = "ADOBECODEBANC", target = "ABC", the minimum window is "BANC"

Challenge

Can you do it in time complexity O(n) ?

Solution

public class Solution {
    public String minWindow(String source, String target) {
        int[] S = new int[255], T = new int[255];
        for (int i = 0; i < target.length(); i++) T[target.charAt(i)]++;
        int start = 0, left = -1, right = source.length(), match = 0, len = source.length();
        for (int i = 0; i < source.length(); i++) {
            S[source.charAt(i)]++;
            if (S[source.charAt(i)] <= T[source.charAt(i)]) match++;
            if (match == target.length()) {
                while (start < i && S[source.charAt(start)] > T[source.charAt(start)]) {
                    S[source.charAt(start)]--;
                    start++;
                }
                if (i - start < len) {
                    len = i - start;
                    left = start;
                    right = i;
                }
                S[source.charAt(start)]--;
                match--;
                start++;
            }
        }
        return left == -1 ? "" : source.substring(left, right+1);
    }
}

Solution: Updated 2018-10

class Solution {
    public String minWindow(String s, String t) {
        if (s == null || t == null || s.length() < t.length()) return "";
        int[] dict = new int[256];
        for (char ch: t.toCharArray()) {
            dict[ch]++;
        }
        int l = 0, r = 0, start = -1;
        int count = t.length();
        int min = Integer.MAX_VALUE;
        while (r < s.length()) {
            char ch = s.charAt(r);
            if (dict[ch] > 0) count--;
            dict[ch]--;
            r++;
            
            while (count == 0) {
                ch = s.charAt(l);
                if (dict[ch] == 0) count++;
                dict[ch]++;
                if (r-l < min) {
                    min = r-l;
                    start = l;
                }
                l++;
            }
        }
        return start == -1 ? "" : s.substring(start, start+min);
    }
}

GPU云服务器云服务器 Minimum Substring SUBSTRING_INDEX java的substring方法

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