LeetCode Brushing Notes _76. Minimum Coverage Substring

The topic is from LeetCode

76. Minimum Covering Substring

Other solutions or source code can be accessed: tongji4m3

description

Give you a string S and a string T. Please design an algorithm that can find out from the string S within O(n) time complexity: the smallest substring containing all characters of T.

Example:

输入：S = "ADOBECODEBANC", T = "ABC"
输出："BANC"

prompt:

如果 S 中不存这样的子串，则返回空字符串 ""。
如果 S 中存在这样的子串，我们保证它是唯一的答案。

Ideas

The method of sliding window: first expand hi to include T, and then increase lo to minimize it until it does not meet the requirements. Continue to loop to expand hi

//初步思路
int lo=0,hi=0;//左右指针,指向最小覆盖子串的首尾
int result=0;
while(hi<N) 
{	
	while([lo,hi]区间包含T)
	{
		result=max(result,hi-lo+1);
		++lo;//增加lo
	}
	++hi;//扩大hi
}

The key is how to judge: [lo,hi] interval contains T

You can make the map store the characters in T and the number of occurrences. If characters in T appear in the loop, they are added to the window

Use a value to determine the match between the map and the characters in Windows (if a character appears the same number of times, it will match)

int lo=0,hi=0;//左右指针,指向最小覆盖子串的首尾
int match=0;//匹配情况

for ch in T:
	map.put(ch,map.get(ch)+1)

while(hi<N) 
{
	char ch=T[hi]
	if(map.contains(ch))
	{
		window.put(ch,window.get(ch)+1);
		if(map.get(ch)==window.get(ch)) ++match; //windows中该字符太多不会匹配多次
	}
	//说明至少包含了T所有字符
	while(match==map.size())
	{
		result=max(result,hi-lo+1);
		if(map.contains(T[lo])) //说明滑动了一个T中字符
		{
			window.put(T[lo],window.get(T[lo])-1);//look ch代表的是T[hi]
			if(map.get(T[lo])>window.get(T[lo])) --match;
		}
		++lo;//增加lo
	}
	++hi;//扩大hi
}

Code

public String minWindow(String s, String t)
{
    
    
    int lo=0,hi=0;//左右指针,指向最小覆盖子串的首尾
    int start=0,length=Integer.MAX_VALUE;//look 记录结果 求最小,length应该为最大
    int match = 0;//匹配情况

    Map<Character, Integer> map = new HashMap<>();
    Map<Character, Integer> window = new HashMap<>();//代表滑动窗口
    for (char ch : t.toCharArray())
    {
    
    
        map.put(ch, map.getOrDefault(ch, 0) + 1);
    }

    while(hi<s.length())
    {
    
    
        char chHi = s.charAt(hi);//look 命名为ch可能导致下面误导用ch表示S[lo]
        if(map.containsKey(chHi)) //如果出现了T中字符,则放入
        {
    
    
            window.put(chHi, window.getOrDefault(chHi, 0) + 1);
            if(map.get(chHi).equals(window.get(chHi))) ++match; //windows中该字符太多不会匹配多次
        }
        while(match==map.size())
        {
    
    
            if(length>hi-lo+1)
            {
    
    
                length = hi - lo + 1;
                start = lo;
            }
            char chLo = s.charAt(lo);
            if(map.containsKey(chLo)) //说明滑动了一个T中字符
            {
    
    
                window.put(chLo,window.get(chLo)-1);
                if(map.get(chLo)>window.get(chLo)) --match;
            }
            ++lo;//增加lo
        }

        ++hi;//扩大hi
    }
    //如果不包含这样的子串,那length没变
    if(length==Integer.MAX_VALUE) return "";
    return s.substring(start,start+length);
}