LeetCode 滑动窗口题型整理

一、滑动窗口题型模板

  

	/*
	 *  滑动窗口类型: 模板
	 */
    public List<Integer> slideWindowMode(String s, String t) {
    	// 1 根据题目返回类型定义数据结构
    	ArrayList<Integer> result = new ArrayList<>();
    	if(t.length()> s.length())
    		return result;
    	// 2 新建Map, Key=字符,value=频率
    	HashMap<Character, Integer> map = new HashMap<>();
    	for(char c: t.toCharArray())
    		map.put(c, map.getOrDefault(c, 0) + 1);
    	
    	// 3 定义变量
    	int counter = map.size();	//  目标字符串中的不同字符种类
    	int begin = 0, end = 0;	// 窗口起始、结束点
    	int len = Integer.MAX_VALUE;
    	
    	// 4 源字符串开始遍历
    	while(end < s.length()) {
    		// 5 统计每个字符
    		char c = s.charAt(end);
    		if(map.containsKey(c)) {
    			map.put(c, map.get(c) - 1);
    			if(map.get(c) == 0)
    				counter--;
    		}
    		end++;

    		// 6 符合情况!begin~end 之间包含 t 的所有字符
    		while(counter == 0) { // 此时 Map 的 value 全部 <= 0
    			char tmpc = s.charAt(begin);
    			if(map.containsKey(tmpc)) {
    				map.put(tmpc, map.get(tmpc) + 1);
    				if(map.get(tmpc) > 0)
    					counter++;
    			}
    			
    			// 7 do sth
    			
    			begin++;
    		}
    	}
    	// 8
    	return result;
    }

  

二、LeetCode 中 几个滑动窗口套用模板实现

  

  1、 76. Minimum Window Substring 

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

    public String minWindow(String s, String t) {
   
		if(t.length()> s.length()) return "";
		
		HashMap<Character, Integer> map = new HashMap<>();
		for(char c: t.toCharArray())
			map.put(c, map.getOrDefault(c, 0) + 1);
		
    	int counter = map.size();	//  目标字符串中的不同字符种类
    	int begin = 0, end = 0;	// 窗口起始、结束点
    	int len = Integer.MAX_VALUE;
    	int head = 0;

    	while(end < s.length()) {
    		char c = s.charAt(end);
    		if( map.containsKey(c) ) {
    			map.put(c, map.get(c) - 1);
    			if(map.get(c) == 0)
    				counter--;
    		}
    		end++;
    		
    		while(counter == 0) {	// 此时 Map 的 value 全部 <= 0
    			char tmpc = s.charAt(begin);
    			if( map.containsKey(tmpc) ) {
    				map.put(tmpc, map.get(tmpc) + 1);
    				if(map.get(tmpc) > 0)
    					counter++;
    			}
    			//----------
    			if(end - begin < len) {
    				len = end - begin;
    				head = begin;
    			}
    			//----------
    			begin++;	
    		}
    	}

    	if(len == Integer.MAX_VALUE )
    		return "";
    	return s.substring(head, head + len);
    }

  

  2、3 Longest Substring Without Repeating Characters

  https://leetcode.com/problems/longest-substring-without-repeating-characters/description/

	/*
	 * 2、 Longest Substring Without Repeating Characters
	 */
    public int lengthOfLongestSubstring(String s) {
    	
    	HashMap<Character, Integer> map = new HashMap<>();
    	int begin = 0;	// 窗口起始、结束点
    	int max = 0;
    	for (int i = 0; i < s.length(); i++) {
    		char c = s.charAt(i);
    		if(map.containsKey(c)) {
    			max = Math.max(max, i - begin);
    			begin = Math.max(begin, map.get(c) + 1);
    			map.put(c, i);
    		}
    		else {
    			map.put(c, i);
    		}
		}
    	max = Math.max(s.length() - begin, max);
    	return max;
    }
	

 

  3、438. Find All Anagrams in a String

  https://leetcode.com/problems/find-all-anagrams-in-a-string/description/

    public List<Integer> findAnagrams(String s, String t) {
        ArrayList<Integer> result = new ArrayList<>();
		if(s.length() < t.length())
			return result;
		
		HashMap<Character, Integer> map = new HashMap<>();
		for(char c: t.toCharArray())
			map.put(c, map.getOrDefault(c, 0) + 1);
		
		int counter = map.size();
		
		int begin = 0, end = 0;
//		int head = 0;
//		int len = Integer.MAX_VALUE;
		while(end < s.length()) {
			char c = s.charAt(end);
			if( map.containsKey(c) ) {
				map.put(c, map.get(c) - 1);
				if(map.get(c) == 0)
					counter--;
			}
			end++;
			
			while(counter == 0) {
				char tmpc = s.charAt(begin);
				if( map.containsKey(tmpc) ) {
					map.put(tmpc, map.get(tmpc) + 1);
					if(map.get(tmpc) > 0)
						counter++;
				}
				//--------
				if(end - begin == t.length())
					result.add(begin);
				//--------
				begin++;
			}
		}
		return result;
    }

  

posted @ 2019-07-16 20:21  skillking2  阅读(786)  评论(0编辑  收藏