leetcode--Minimum Window Substring

最新推荐文章于 2021-02-25 13:50:42 发布

转载最新推荐文章于 2021-02-25 13:50:42 发布 · 139 阅读

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原文链接：http://www.cnblogs.com/averillzheng/p/3544185.html

本文深入探讨了如何在字符串S中寻找包含字符串T所有字符的最小子串，提出了复杂度为O(n)的高效算法，并通过三个不同实现展示了如何使用HashMap和数组来优化查找过程。

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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.

code 1

public class Solution {
    public String minWindow(String S, String T) {
       if(T.length() == 0 || S.length() < T.length())
			return "";
		Map<Character, Integer> mult = new HashMap<Character, Integer>();
		Map<Character, Integer> alreadyRead = new HashMap<Character, Integer>();
		for(int i = 0; i < T.length(); ++i){
			if(mult.containsKey(T.charAt(i)))
				mult.put(T.charAt(i), mult.get(T.charAt(i)) + 1);
			else {
				mult.put(T.charAt(i), 1);
				alreadyRead.put(T.charAt(i), 0);
			}
		}
		
		String minWindow = "";
		int cnt = 0, start = 0, end = 0, currentLength = S.length() + 1;
		for(; end < S.length(); ++end){
			if(!mult.containsKey(S.charAt(end)))
				continue;
			else {
				int num = alreadyRead.get(S.charAt(end));
				alreadyRead.put(S.charAt(end), num + 1);
				if(num + 1 <= mult.get(S.charAt(end)))
					++cnt;
			}
			
			if(cnt == T.length()){
				while(!mult.containsKey(S.charAt(start)) || alreadyRead.get(S.charAt(start)) > mult.get(S.charAt(start))){
				    if(mult.containsKey(S.charAt(start))){
					    int num = alreadyRead.get(S.charAt(start));
					    alreadyRead.put(S.charAt(start), num - 1);
				    }
					++start;
				}
				
				if(end - start + 1 < currentLength) {
					minWindow = S.substring(start, end + 1);
					currentLength = end - start + 1;
				}
			}
		}
		return minWindow;
    }
}

the following two codes follow the same algorithm with different data structures. Both are accepted

public class Solution {
    public String minWindow(String S, String T) {
       String result = "";
		int slen = S.length(), tlen = T.length();
		if(slen > 0 && tlen > 0 && slen >= tlen){
			HashMap<Character, Integer> mult = new HashMap<Character, Integer>();
			for(int i = 0; i < tlen; ++i){
				int mu = 1;
				if(mult.containsKey(T.charAt(i)))
					mu = mult.get(T.charAt(i)) + 1;
				mult.put(T.charAt(i), mu);
			}
			//find the substring of S which contains all characters in T
			int size = mult.size();
			int start = 0, end = 0;
			for(int i = 0; i < slen; ++i){
				if(mult.containsKey(S.charAt(i))){
					int mu = mult.get(S.charAt(i)) - 1;
					if(mu == 0)
						--size;
					mult.put(S.charAt(i), mu);
				}
				if(size == 0)
					break;
				++end;
			}
			
			if(size == 0){				
				start = shorten(start, end, mult, S);
				result = S.substring(start, end + 1);
				int len = end - start + 1, newend = end;
				while(newend + 1 < slen){
					++newend;
					if(mult.containsKey(S.charAt(newend))){
						int mu = mult.get(S.charAt(newend)) - 1;
						mult.put(S.charAt(newend), mu);
						start = shorten(start, newend, mult, S);
						if(newend - start + 1 < len){
							result = S.substring(start, newend + 1);
							len = newend - start + 1;
							end = newend;
						}
					}
				}
			}			
		}
		return result;       
    }
	
	public int shorten(int start, int end, HashMap<Character, Integer> mult, String S){
		while(start < end){
			if(!mult.containsKey(S.charAt(start)))
				++start;
			else{
				if(mult.get(S.charAt(start)) < 0){
					int j = mult.get(S.charAt(start)) + 1;
					mult.put(S.charAt(start), j);
					++start;							
				}	
				else
					break;
			}	
		}
		return start;
    }
}

Another code

public class Solution {
    public String minWindow(String S, String T) {
        String result = "";
		int slen = S.length(), tlen = T.length();
		if(slen > 0 && tlen > 0 && slen >= tlen){
			int[] target = new int[256];
			int distChar = 0;
			for(int i = 0; i < tlen; ++i){
				++target[T.charAt(i)];
				distChar = (target[T.charAt(i)] == 1)? distChar + 1 : distChar;
			}
			
			int start = 0, end = 0;
			int[] source = new int[256];			
			for(int i = 0; i < slen; ++i){
				if(target[S.charAt(i)] != 0){
					++source[S.charAt(i)];
					distChar = (source[S.charAt(i)] == target[S.charAt(i)]) ? distChar - 1: distChar; 
				}
				if(distChar == 0)
					break;
				++end;
			}
			if(distChar == 0){
				start = shorten(target, source, start, end, S);
				result = S.substring(start, end + 1);
				int len = end - start + 1, newend = end;
				while(newend + 1 < slen){
					++newend;
					if(target[S.charAt(newend)] != 0){
						++source[S.charAt(newend)];
						start = shorten(target, source, start, newend, S);
						if(newend - start + 1 < len){
							len = newend - start + 1;
							end = newend;
							result = S.substring(start, end + 1);
						}
					}
				}
			}		
		}
		return result;
	}
	
	public static int shorten(int[] target, int[] source, int start, int end, String S){
		while(start < end){
			if(target[S.charAt(start)] == 0)
				++start;
			else{ 
				if(source[S.charAt(start)] > target[S.charAt(start)]){
					--source[S.charAt(start)];
					++start;
				}
				else
					break;
			}			
		}
		return start;
    }
}

转载于:https://www.cnblogs.com/averillzheng/p/3544185.html